Study Notes on Indefinite Integrals

Chapter 5: Integrals - Part 1: Indefinite Integrals

5.1 Antiderivatives and Indefinite Integrals

Definition 1: Let $f : I ightarrow ext{R}$ be a real function defined on an interval $I$ . A function $F : I ightarrow ext{R}$ is called an antiderivative of $f$ on $I$ if:
1. $F$ is differentiable on $I$ , and
2. For every $x ext{ in } I$ , $F^{ ext{'}}(x) = f(x)$ .

Examples of Antiderivatives

Let $f : ext{R} ightarrow ext{R}, f(x) = 2x$ . The functions:
- $F_1(x) = x^2 + 1$
- $F_2(x) = x^2 - 3$
 are two antiderivatives of $f$ on $ext{R}$ .
The functions $x ightarrow ext{sin}(x) + c$ (where $c ext{ in } ext{R}$ ) are antiderivatives of the function $x ightarrow ext{cos}(x)$ on $ext{R}$ .
The functions $x ightarrow ext{ln}(x) + c$ (where $c ext{ in } ext{R}$ ) are antiderivatives of the function $x ightarrow rac{1}{x}$ on the interval $(0,+ ext{∞})$ .
The functions $x ightarrow ext{ln}(-x) + c$ (where $c ext{ in } ext{R}$ ) are antiderivatives of the function $x ightarrow rac{1}{x}$ on the interval $(- ext{∞}, 0)$ .

Proposition 1

If a real function $f$ has an antiderivative $F$ on an interval $I$ , then the antiderivatives of $f$ are the functions $F + c$ , where $c$ is any constant.

Proof: Let $G:I ightarrow ext{R}$ defined by $G(x) = F(x) + c$ , where $c ext{ in } ext{R}$ . Then:
$G^{ ext{'}}(x) = F^{ ext{'}}(x) = f(x) ext{ for all } x ext{ in } I.$ So, $G$ is also an antiderivative of $f$ . Conversely, if $F, G : I ightarrow ext{R}$ are two antiderivatives of the same real function $f : I ightarrow ext{R}$ , then we have:

$(F - G)^{ ext{'}}(x) = F^{ ext{'}}(x) - G^{ ext{'}}(x) = f(x) - f(x) = 0 ext{ for all } x ext{ in } I.$

Since $I$ is an interval, from the relation above it follows that the function $F - G$ is constant on $I$ . Hence, $G(x) = F(x) + c$ for all $x ext{ in } I$ , with $c ext{ in } ext{R}$ .

Remark 1

If $f$ has an antiderivative $F$ on the interval $I$ , then all antiderivatives of $f$ are functions of the form $F + c$ . This implies that $f$ has infinitely many antiderivatives differing by a constant.

Remark 2

Not every function has an antiderivative. However, we will prove later that every continuous function on an interval has an antiderivative on that interval.
There exist discontinuous functions having an antiderivative. For instance, the function $F$ defined on $ext{R}$ by:
F(x) =egin{cases}
x^2 ext{sin}(1/x) & ext{if } x
eq 0 \
0 & ext{if } x = 0
\ ext{is an antiderivative of the function} \ f ext{ defined on } ext{R} ext{ by } f(x) = egin{cases}
2x ext{sin}(1/x) - ext{cos}(1/x) & ext{if } x
eq 0 \
0 & ext{if } x = 0
\

which is discontinuous at zero.

5.2 General Methods for Computing Indefinite Integrals

Remarks and Notations

If F $is an antiderivative of$ f $on interval$ I $, we write: $ \int f(x) \, dx = F(x) + c , c ext{ in } ext{R} ag{5.1} $. The notation$ \int f(x) \, dx $is called the indefinite integral of$ f $.</li><li>The variable$ x $in the relation (5.1) is called the variable of integration. Moreover, the symbol$ dx $in (5.1) indicates the independent variable$ x $with respect to which the original differentiation was made. It also suggests that the integration process has to be performed with respect to$ x $.</li><li>The constant$ c $in the relation (5.1) is called the constant of integration.</li><li>The function$ f(x) $in the relation (5.1) is called the integrand.</li><li>The question "calculate an indefinite integral$ \int f(x) \, dx $" means: Find an antiderivative of$ f $, defined up to an additive constant.</li></ol><h5 id="b7cde960-81d7-4b7e-98d3-f503342fd7f2" data-toc-id="b7cde960-81d7-4b7e-98d3-f503342fd7f2" collapsed="false" seolevelmigrated="true">Examples of Indefinite Integrals</h5><ol><li>Since$ x
ightarrow x^2 $is an antiderivative of$ x
ightarrow 2x $on$ ext{R} $, we write: $ \int 2x \, dx = x^2 + c , c ext{ in } ext{R} $.</li><li>Since$ x
ightarrow ext{sin}(x) $is an antiderivative of$ x
ightarrow ext{cos}(x) $on$ ext{R} $, we write: $ \int ext{cos}(x) \, dx = ext{sin}(x) + c , c ext{ in } ext{R} $.</li></ol><h5 id="8331827d-278d-431f-807a-efd1ed9b8421" data-toc-id="8331827d-278d-431f-807a-efd1ed9b8421" collapsed="false" seolevelmigrated="true">Remark 3</h5><ul><li>The question "prove that the function$ F $is an antiderivative of$ f $on an interval$ I $" is equivalent to the question "prove that for all$ x ext{ in } I: \int f(x) \, dx = F(x) + c , c ext{ in } ext{R} $. To answer this question, it suffices to show that$ F^{ ext{'}}(x) = f(x) $on$ I $.</li></ul><h5 id="c7c795ca-5058-4d47-b387-efa51113de06" data-toc-id="c7c795ca-5058-4d47-b387-efa51113de06" collapsed="false" seolevelmigrated="true">Example 2</h5><ul><li>Prove that for all$ x ext{ in } ext{R} $we have: $ \int \frac{x}{x^2+x+1} \, dx = rac{1}{2} ext{ln}(x^2 + x + 1) - rac{1}{3} ext{arctan}igg( rac{2x + 1}{ ext{sqrt{3}}}igg) + c , c ext{ in } ext{R}. $</li></ul><h5 id="da3fd918-7d7c-4151-a5c6-6017e7d1fbef" data-toc-id="da3fd918-7d7c-4151-a5c6-6017e7d1fbef" collapsed="false" seolevelmigrated="true">5.2.1 Antiderivatives of Common Functions</h5><ul><li>The table below lists the antiderivatives of common functions, which are determined by reversing the process of differentiation and are essential for future calculations of integrals (both antiderivatives and definite integrals).</li></ul><table style="min-width: 75px;"><colgroup><col style="min-width: 25px;"><col style="min-width: 25px;"><col style="min-width: 25px;"></colgroup><tbody><tr><th colspan="1" rowspan="1">Function$ f(x) $</th><th colspan="1" rowspan="1">Antiderivative$ \int f(x) \, dx $</th><th colspan="1" rowspan="1">Domain</th></tr><tr><td colspan="1" rowspan="1">$ x^n, n ext{ in } ext{R} $</td><td colspan="1" rowspan="1">$ \frac{x^{n+1}}{n+1} + c $</td><td colspan="1" rowspan="1">All$ x $except where$ n=-1 $</td></tr><tr><td colspan="1" rowspan="1">$ \frac{1}{x^m}, m ext{ in } ext{R}, m
eq 1 $</td><td colspan="1" rowspan="1">$ \frac{x^{-m+1}}{-m+1} + c = -\frac{1}{m-1} x^{m-1} + c $</td><td colspan="1" rowspan="1">All$ x $,$ m
eq 0 $</td><td colspan="1" rowspan="1"></td><td colspan="1" rowspan="1"></td></tr><tr><td colspan="1" rowspan="1">$ e^x $</td><td colspan="1" rowspan="1">$ e^x + c $</td><td colspan="1" rowspan="1">All$ x $</td></tr><tr><td colspan="1" rowspan="1">$ ext{sinh}(x) $</td><td colspan="1" rowspan="1">$ ext{cosh}(x) + c $</td><td colspan="1" rowspan="1">All$ x $</td></tr><tr><td colspan="1" rowspan="1">$ ext{cosh}(x) $</td><td colspan="1" rowspan="1">$ ext{sinh}(x) + c $</td><td colspan="1" rowspan="1">All$ x $</td></tr><tr><td colspan="1" rowspan="1">$ ext{cos}(x) $</td><td colspan="1" rowspan="1">$ ext{sin}(x) + c $</td><td colspan="1" rowspan="1">All$ x $</td></tr><tr><td colspan="1" rowspan="1">$ ext{sin}(x) $</td><td colspan="1" rowspan="1">$ - ext{cos}(x) + c $</td><td colspan="1" rowspan="1">All$ x $</td></tr><tr><td colspan="1" rowspan="1">$ rac{1}{ ext{cos}^2(x)} $</td><td colspan="1" rowspan="1">$ ext{tan}(x) + c $</td><td colspan="1" rowspan="1">All$ x $except odd multiples of$ \frac{ ext{π}}{2} $</td></tr><tr><td colspan="1" rowspan="1">$ rac{1}{ ext{sin}^2(x)} $</td><td colspan="1" rowspan="1">$ - ext{cot}(x) + c $</td><td colspan="1" rowspan="1">All$ x $except multiples of$ ext{π} $</td></tr><tr><td colspan="1" rowspan="1">$ rac{1}{x^2+1} $</td><td colspan="1" rowspan="1">$ ext{arctan}(x) + c $</td><td colspan="1" rowspan="1">All$ x $</td></tr><tr><td colspan="1" rowspan="1">$ rac{1}{ ext{sqrt{1-x^2}}} $</td><td colspan="1" rowspan="1">$ ext{arcsin}(x) + c $</td><td colspan="1" rowspan="1">$ -1 < x < 1 $</td></tr><tr><td colspan="1" rowspan="1">$ rac{1}{ ext{cosh}^2(x)} $</td><td colspan="1" rowspan="1">$ ext{argsinh}(x) + c = ext{ln}(x + ext{sqrt{x}^2+1}) + c $</td><td colspan="1" rowspan="1">All$ x $</td></tr><tr><td colspan="1" rowspan="1">$ rac{1}{ ext{sqrt{x^2-1}}} $</td><td colspan="1" rowspan="1">$ ext{argch}(x) + c = ext{ln}(x + ext{sqrt{x}^2-1}) + c $</td><td colspan="1" rowspan="1">All$ x, x > 1 $</td></tr></tbody></table><h5 id="5aa0beb4-d905-4685-a151-a403e56b64b4" data-toc-id="5aa0beb4-d905-4685-a151-a403e56b64b4" collapsed="false" seolevelmigrated="true">5.2.2 Operations on Antiderivatives</h5><ul><li>By reversing the process of differentiation, we can show:</li></ul>Proposition 2: If$ f $and$ g $are two real functions having antiderivatives$ F $and$ G $on a certain interval$ I $, then for all$ x ext{ in } I $and for all$ ext{α, β in } ext{R} $, we have: $ \int (\alpha f(x) + \beta g(x)) \, dx = \alpha \int f(x) \, dx + \beta \int g(x) \, dx = \alpha F(x) + \beta G(x) + c $.<h5 id="7a8d0c5d-e55d-4206-a36f-67b436b90a32" data-toc-id="7a8d0c5d-e55d-4206-a36f-67b436b90a32" collapsed="false" seolevelmigrated="true">Example 2</h5><ul><li>Calculate the following integrals:<ol><li>$ \int (x^4 - 2x^3 - 1) \, dx $</li><li>$ \int (2 ext{sin}(x) + 5 ext{cos}(x)) \, dx $</li><li>$ \int \frac{4}{x^2 + 1} \, dx $</li></ol></li></ul>Answers:<ol><li>Using the linearity of the integral, we see that: $ \int (x^4 - 2x^3 - 1) \, dx = \frac{x^5}{5} - \frac{x^4}{2} - x + c, c ext{ in } ext{R} $.</li><li>We can easily find: $ \int (2 ext{sin}(x) + 5 ext{cos}(x)) \, dx = 2\int ext{sin}(x) \, dx + 5\int ext{cos}(x) \, dx = -2 ext{cos}(x) + 5 ext{sin}(x) + c, c ext{ in } ext{R} $.</li><li>$ \int \frac{4}{x^2 + 1} \, dx = 4\int \frac{1}{x^2 + 1} \, dx = 4 ext{arctan}(x) + c, c ext{ in } ext{R} $.</li></ol><h5 id="e2c8a304-4d80-49a3-bd5c-9190cee540b1" data-toc-id="e2c8a304-4d80-49a3-bd5c-9190cee540b1" collapsed="false" seolevelmigrated="true">Remark 4</h5><ul><li>We can integrate any function that is the derivative of a known function, but there are elementary functions that are not the derivative of any elementary function, such as:<ul><li>$ e^{-x^{2}} $,$ ext{sin}(x) ext{ln}(x) $, and$ e^{x} x $. This means that the antiderivative is a new function, not belonging to a catalog of well-known elementary functions.</li></ul></li></ul><h5 id="81aad9c0-6aab-4095-9206-48aba2216037" data-toc-id="81aad9c0-6aab-4095-9206-48aba2216037" collapsed="false" seolevelmigrated="true">5.2.3 Integration by Parts for Indefinite Integrals</h5>Theorem 1: Let$ u $and$ v $be two functions continuously differentiable on an interval$ I $. We have: $ \int u^{ ext{'}}(x)v(x) \, dx = u(x)v(x) - \int u(x)v^{ ext{'}}(x) \, dx ag{5.2} $.Proof: We have: $ (uv)^{ ext{'}} = u^{ ext{'}}v + uv^{ ext{'}}\Rightarrow u^{ ext{'}}v = (uv)^{ ext{'}} - uv^{ ext{'}} \Rightarrow \int u^{ ext{'}}(x)v(x) \, dx = \int (uv)^{ ext{'}}(x) \, dx - \int (uv^{ ext{'}})(x) = u(x)v(x) - \int u(x)v^{ ext{'}}(x) \, dx $.<h5 id="1fa9bbc3-f611-40a6-8178-6ceecf4d37a8" data-toc-id="1fa9bbc3-f611-40a6-8178-6ceecf4d37a8" collapsed="false" seolevelmigrated="true">Example 3</h5><ul><li>Calculate:$ \int x ext{sin}(x) \, dx $. Answer: We use integration by parts (IBP). Let:<ul><li>$ u = x $</li><li>$ v^{ ext{'}} = ext{sin}(x) $\Rightarrow$ v = - ext{cos}(x) \
Conversion:
\int x ext{sin}(x) \, dx = -x ext{cos}(x) + \int - ext{cos}(x) \, dx = -x ext{cos}(x) + ext{sin}(x) + c, c ext{ in } ext{R} $.</li></ul></li></ul><h5 id="2fe77cac-c8e7-4036-b5ad-6a8bafd8d0a5" data-toc-id="2fe77cac-c8e7-4036-b5ad-6a8bafd8d0a5" collapsed="false" seolevelmigrated="true">5.2.4 Substitution Method for Indefinite Integrals</h5>Proposition 3 (First Substitution Method): Let I and J be two intervals of$ ext{R} $,$ ig[ ext{implies} $,$ ext{ψ} : I → J $a differentiable function, and$ f : J
ightarrow ext{R} $a continuous function. If$ F $is an antiderivative of$ f $on$ J $, then$ F ig ext{◦} ext{ψ} $is an antiderivative of$ (f ig ext{◦} ext{ψ}) imes ext{ψ}^{ ext{'}} $on$ I $.<h6 id="bf0d8e20-8c5c-4b71-86d8-5ca10f8aa80c" data-toc-id="bf0d8e20-8c5c-4b71-86d8-5ca10f8aa80c" collapsed="false" seolevelmigrated="true">Indefinite Integral Representation</h6>$ \int f(\text{ψ}(x)) ext{ψ}^{ ext{'}}(x) \, dx = \int f(t) \, dt = F(t) + c = F(\text{ψ}(x)) + c $, where$ t = ext{ψ}(x) $.Proof: Since$ F $is an antiderivative of$ f $on$ J $, it is differentiable on$ J $and its derivative is$ f $. Since$ ext{ψ} $is differentiable on$ I $, the composite function$ F ig ext{◦} ext{ψ} $is differentiable on$ I $and its derivative is: $ (F ig ext{◦} ext{ψ})^{ ext{'}}(x) = F^{ ext{'}}( ext{ψ}(x)) ext{ψ}^{ ext{'}}(x) = f( ext{ψ}(x)) ext{ψ}^{ ext{'}}(x). $ This means that$ F ig ext{◦} ext{ψ} $is an antiderivative of$ (f ig ext{◦} ext{ψ}) imes ext{ψ}^{ ext{'}} $on$ I $.<h6 id="c28b1b08-3653-4077-b714-60c3058e24b8" data-toc-id="c28b1b08-3653-4077-b714-60c3058e24b8" collapsed="false" seolevelmigrated="true">First Substitution Method Steps</h6><ol><li>Set$ t = ext{ψ}(x) $(the substitution or change of variable, from$ x $to$ t $using the rule$ t = ext{ψ}(x) $).</li><li>$ rac{dt}{dx} = ext{ψ}^{ ext{'}}(x) \Rightarrow dt = ext{ψ}^{ ext{'}}(x) \, dx $so the expression$ f( ext{ψ}(x)) ext{ψ}^{ ext{'}}(x) \, dx $is replaced by$ f(t) \, dt $.</li><li>Compute$ \int f(t) \, dt = F(t) + c $.</li><li>Return to the original variable$ x $, replace$ t = ext{ψ}(x) $, we obtain: $ \int f( ext{ψ}(x)) ext{ψ}^{ ext{'}}(x) \, dx = \int f(t) \, dt = F(t) + c = F( ext{ψ}(x)) + c, c ext{ in } ext{R}. $</li></ol><h5 id="e1089347-41be-4ec7-91bf-4994ae176e72" data-toc-id="e1089347-41be-4ec7-91bf-4994ae176e72" collapsed="false" seolevelmigrated="true">Example 4</h5><ul><li>Compute: $ \int e^{x} e^{2x + 1} \, dx $. Answer:</li></ul><ol><li>Let$ t = e^{x}, x ext{ in } ext{R} $. Thus$ dt = e^{x} \, dx $. Hence: $ \int e^{x}e^{2x+1} \, dx = \int \frac{1}{t^2} + dt = ext{arctan}(t) + c = ext{arctan}(e^{x}) + c, c ext{ in } ext{R}. $</li></ol><h5 id="8f9b946b-0049-4327-9caf-d822fddb115f" data-toc-id="8f9b946b-0049-4327-9caf-d822fddb115f" collapsed="false" seolevelmigrated="true">Example 5</h5><ul><li>Compute: $ \int (\ln x)^{2022} x \, dx. $</li></ul><ol><li>Take$ t = ext{ln} x, x > 0. $Thus$ dt = \frac{1}{x} \, dx $.</li></ol>Overall solution: $ \int (\ln x)^{2022} x \, dx = \int t^{2022} dt = \frac{t^{2023}}{2023} + c = \frac{(\ln x)^{2023}}{2023} + c, c ext{ in } ext{R}. $<h5 id="4a432d76-7f9a-4409-9eaf-9a0a50779377" data-toc-id="4a432d76-7f9a-4409-9eaf-9a0a50779377" collapsed="false" seolevelmigrated="true">5.2.5 Applications</h5>Application 1: Proposition 2<ol><li>$ \int u^{ ext{'}}(x) ext{cos}(u(x)) \, dx = ext{sin}(u(x)) + c, x ext{ in } I, c ext{ in } ext{R}. $</li><li>$ \int u^{ ext{'}}(x) ext{sin}(u(x)) \, dx = - ext{cos}(u(x)) + c, x ext{ in } I, c ext{ in } ext{R}. $</li><li>$ \int u^{ ext{'}}(x)u(x) \, dx = ext{ln}|u(x)| + c, ext{ if } u(x)
eq 0 ext{ in } I, c ext{ in } ext{R}. $</li><li>$ \int u^{ ext{'}}(x)(u(x))^{ ext{α}} \, dx = \frac{1}{ ext{α}+1}(u(x))^{ ext{α}+1} + c, \text{ if } u(x) > 0 ext{ in } I, ext{α} ext{ in } ext{R}, ext{α}
eq -1, c ext{ in } ext{R}. $</li></ol>Proof: In each case, change variable$ t = u(x) $and then use antiderivatives of standard functions.Application 2: Proposition 3<ol><li>$ \int \frac{1}{x^{2}+a^{2}} \, dx = \frac{1}{a} ext{arctan}\left( rac{x}{a}\right) + c, a > 0, c ext{ in } ext{R}, x ext{ in } ext{R}. $</li><li>$ \int \frac{1}{x^{2}-a^{2}} \, dx = \frac{1}{2a} ext{ln}\left|\frac{x-a}{x+a}\right| + c, a
eq 0, c ext{ in } ext{R}, x
eq ext{±} a. $</li><li>$ \int \frac{1}{ ext{a}^2-x^2} \, dx = ext{arcsin}\left( rac{x}{a}\right) + c, a > 0, c ext{ in } ext{R}, |x| < a. $</li><li>$ \int \frac{1}{ ext{a}^2 + x^2} \, dx = ext{argsinh}\left(\frac{x}{a}\right) + c = ext{ln}\left(x + ext{sqrt}(x^{2} + a^{2}) \right) + c, a > 0, c ext{ in } ext{R}, x ext{ in } ext{R}. $</li><li>$ \int \frac{1}{−x^{2}+a^{2}} \, dx = ext{argcosh}\left(\frac{x}{a}\right) + c = ext{ln}\left(x + ext{sqrt}(x^{2} - a^{2})\right) + c, a > 0, c ext{ in } ext{R}, x > a. $</li></ol>Proof: In each case, make the substitution$ t = rac{u}{a} $, and then use the antiderivatives of standard functions.<h5 id="57fbcad4-daca-406d-9147-1dab26130bea" data-toc-id="57fbcad4-daca-406d-9147-1dab26130bea" collapsed="false" seolevelmigrated="true">Remark 6</h5><ul><li>In computing indefinite integrals (i.e., antiderivatives), we utilize the results of Propositions 2 and 3 directly. For example: $ \int \frac{1}{x^{2} + 3} \, dx = \frac{1}{3} ext{arctan}\left(\frac{x}{\sqrt{3}}\right) + c, c ext{ in } ext{R}. $</li></ul><h4 id="017ad7ab-c605-48ce-ab04-d7bfb6a8252a" data-toc-id="017ad7ab-c605-48ce-ab04-d7bfb6a8252a" collapsed="false" seolevelmigrated="true">5.3 Integration of Certain Expressions Containing the Trinomial$ ax^2 + bx + c $</h4><h5 id="f614ee21-4685-48da-973b-2c68602759a8" data-toc-id="f614ee21-4685-48da-973b-2c68602759a8" collapsed="false" seolevelmigrated="true">5.3.1 Evaluation of integral$ I(x) = \int A x + B \frac{1}{ax^2 + bx + c} \, dx $</h5><ul><li>Method: According to the sign of$ ext{Δ} = b^2 - 4ac $. There are three cases to consider:</li></ul><ol><li>Case$ ext{Δ} > 0 $:<ul><li>In this case,$ P(x) $has two distinct real roots$ x_1 < x_2 $and we have$ P(x) = a(x - x_1)(x - x_2) $. We use partial fraction decomposition: $ A x + B = \frac{p}{x - x_1} + \frac{q}{x - x_2}. $ Here$ p $and$ q $are two real numbers that need to be determined. Thus for all$ x ext{ in } ext{R} $except$ ext{x} = x_1, x_2 $, we have: $ I(x) = \int (Ax + B) \frac{1}{a(x - x_1)(x - x_2)} \, dx = \int \left( \frac{p}{x - x_1} + \frac{q}{x - x_2} \right) \,, dx = p ext{ln}|x - x_1| + q ext{ln}|x - x_2| + c, c ext{ in } ext{R}. $</li></ul></li><li>Case$ ext{Δ} = 0 $:<ul><li>Here,$ P(x) $has a double root$ x_0 = -\frac{b}{2a} $and we denote$ P(x) = a(x - x_0)^2 $. Thus for all$ x ext{ in } ext{R} $except$ x = x_0 $we have: $ I(x) = \int \frac{A x + B}{a(x - x_0)^2} \, dx = \frac{1}{a} \left[ A(x - x_0) + (B + A x_0)(x - x_0)^2 \right], c ext{ in } R. $</li></ul></li><li>Case$ ext{Δ} < 0 $:<ul><li>In this case, we consider the trinomial$ P(x) = a x^2+bx+c $with no real roots. In this case, we will apply integration via substitution: Set$ t = x + \frac{b}{2a} $, then$ d x = dt $, and we obtain: $ I(x) = \int \frac{1}{a(t^2 + k^2)} \, dt. $ Thus we can use trigonometric identities to derive an expression for the integral: $ I(x) = \frac{1}{a} \text{argucoth}\left(\frac{t}{k}\right) + c. $</li></ul></li></ol>Example 6: Calculate:<ul><li>$ G_1(x) = \int \frac{x-3}{x^2-3x+2} \, dx\ $</li><li>$ G_2(x) = \int \frac{5x+1}{x^2-4x+4} \, dx\ $</li><li>$ G_3(x) = \int \frac{x-3}{x^2+x+1} \, dx. $</li></ul><h5 id="c2fc4cee-5186-4a46-9726-8cc17834f297" data-toc-id="c2fc4cee-5186-4a46-9726-8cc17834f297" collapsed="false" seolevelmigrated="true">Answers:</h5><ol><li>Calculation of$ G_1(x) $:</li></ol><ul><li>The trinomial is factorable:$ x^2 - 3x + 2 = (x - 1)(x - 2) $, with respective roots$ x = 1 $and$ x = 2 $. Thus we can perform partial fraction decomposition, giving: $ g(x) = a rac{1}{x-1} + b rac{1}{x-2} $. Thus, we have: $ G_1(x) = 2 ext{ln}|x - 1| - ext{ln}|x - 2| + c, c ext{ in } ext{R}. $</li></ul><ol><li>Start Calculation for$ G_2(x) $: $ G_2 = 5 \int rac{1}{(x - 2)^2} + 11\ int rac{1}{(x - 2)^2} = 5[ ext{ln}|x - 2| - 11]] \text{from the decomposition received from the comparison of the divisors.} $</li><li>Start Calculation for$ G_3(x) $: $ G_3 = 12 ((x+1)^2) $$; since we have a negative discriminant we cannot deduce any useful //////////////////