Integrals/Antiderivatives

$$\int x^{n}\,dx\ (n\ne-1)$$

$\frac{x^{n+1}}{n+1} + C$; Why can n not equal to -1?

Answer: Recall that this antiderivative comes from the “inverse” power rule. This would require us to add 1 to the power and divide by the new power, which results in: $\frac{x^0}{0}\Rightarrow\text{undefined}$. Alternatively, when $\left(n\ne-1\right)$, we would instead use the absolute-value-natural-logarithm antiderivative of $\frac{1}{x}$.

$$\int \frac{1}{x} \, dx$$

$\ln|x| + C$; Why do we take the absolute value of “x”?

Answer: The domain of $\frac{1}{x}\,$ includes x<1; however, the antiderivative of it cannot. Therefore, the absolute value restricts the argument of the antiderivative, $\ln|x| + C$, to not make an undefined statement.

$$\int e^x \, dx$$

$$e^x + C$$

$$\int a^x \, dx$$

$$\frac{a^x}{\ln a} + C$$

$$\int \cos x \, dx$$

$$\sin x + C$$

$$\int \sin x \, dx$$

$$-\cos x + C$$

$$\int \sec^2 x \, dx$$

$$\tan x + C$$

$$\int \csc^2 x \, dx$$

$$-\cot x + C$$

$$\int \sec x \tan x \, dx$$

$$\sec x + C$$

$$\int \csc x \cot x \, dx$$

$$-\csc x + C$$

$$\int \frac{1}{\sqrt{1 - x^2}} \, dx$$

$$\arcsin x + C$$

$$\int \frac{-1}{\sqrt{1 - x^2}} \, dx$$

$$\arccos x + C$$

$$\int \frac{1}{1 + x^2} \, dx$$

$$\arctan x + C$$

$$\int \frac{1}{a^2 + x^2} \, dx$$

$$\frac{1}{a}\arctan\left(\frac{x}{a}\right) + C$$

$$\int \frac{1}{x^2 - a^2} \, dx$$

$$\frac{1}{2a}\ln\left|\frac{x - a}{x + a}\right| + C$$

$$\int \ln x \, dx$$

$$x \ln x - x + C$$

$$\int e^{kx} \, dx$$

$$\frac{1}{k} e^{kx} + C$$

$$\int a^{kx} \, dx$$

$$\frac{a^{kx}}{k \ln a} + C$$

$$\int f^{\prime}(x)e^{\left\lbrack f(x)\right\rbrack}\,dx$$

$$e^{f(x)} + C$$

$$\int \frac{f'(x)}{f(x)} \, dx$$

$\ln|f(x)| + C$ ; Why do we take the absolute value of “$f(x)$”?

Answer: The domain of $\frac{f'(x)}{f(x)}\,$includes x<1; however, the antiderivative of it cannot. Therefore, we require the absolute value to restrict the argument of the antiderivative, $\ln|f(x)|+C$, to not cause an undefined statement.

$$\int f'(x)\cos(f(x)) \, dx$$

$$\sin(f(x)) + C$$

$$\int f'(x)\sin(f(x)) \, dx$$

$$-\cos(f(x)) + C$$

By parts?

$$\int f^{}\left(x\right)g^{\prime}\left(x\right)=f\left(x\right)\,g\left(x\right)-\int^{}f^{\prime}\left(x\right)g\left(x\right)$$

What is the “$+C$” term in certain integrals and when/why do we include it in certain integrals?

The “constant of integration” is added to indefinite integrals because differentiation maps multiple functions, differing by a constant, to the same derivative. It represents a family of possible antiderivative functions.