Calc 2 Exam 1

∫(2x-3)^-7dx

−1/_{[12⁢(2⁢𝑥−3)6]} +c

∫x/_√(x²+1)

√(𝑥²+1)+𝐶

∫cos³⁢𝜃⁢sin⁡𝜃⁢𝑑⁢𝜃

−cos⁴⁢𝜃/₄+𝐶

∫𝑥²/_{(𝑥³−3)²}𝑑𝑥

−1/ 3(𝑥³−3) + C

∫₀¹ 𝑥/_{√(1+𝑥²)}𝑑𝑥

[1/2]ln|√(1+x²)+x|

∫₀¹ 𝑡²/_{√(1+𝑡³)}𝑑𝑡

2/3(√2−1)

∫₀^π/4 sin⁡𝜃/_{cos⁴⁢𝜃}𝑑𝜃

1/3⁢(2⁢√2−1)

∫₀^√π 𝑡⁢cos⁡(𝑡²)⁢sin⁡(𝑡²)𝑑𝑡

𝑢=sin⁡(𝑡²); the integral becomes 1/2⁢∫₀⁰𝑢𝑑𝑢.

∫3^−𝑥⁢𝑑⁢𝑥

−3^−𝑥/_ln⁡3+𝐶

∫𝑑𝑥/_{𝑥⁢(ln⁡𝑥)²}

−1/_ln⁡𝑥+𝐶

∫(cos⁡𝑥−𝑥⁢sin⁡𝑥)/_{𝑥⁢cos⁡𝑥}𝑑𝑥

ln⁡(𝑥⁢cos⁡𝑥)+𝐶

∫𝑥²⁢𝑒^−𝑥3⁢𝑑⁢𝑥

(−𝑒^{−𝑥^3})/₃+𝐶

∫𝑒^tan⁡𝑥⁢sec²⁢𝑥⁢𝑑⁢𝑥

𝑒^tan⁡𝑥+𝐶

∫₁² 1+2⁢𝑥+𝑥²/_{(3⁢𝑥+3⁢𝑥2+𝑥3)}𝑑𝑥

1/3⁢ln⁡(26/7)

∫_𝜋/3^𝜋/4 cot⁡𝑥⁢𝑑⁢𝑥

1/2⁢ln⁡(3/2)

Integrate: 𝑓⁡(𝑥)=2^−𝑥 𝑎=3 𝑏=4

¹/_{ln⁡(65,536)}

∫₀^√3/2 𝑑𝑥/_{√(1−𝑥2)}

π/3

∫ 𝑑𝑥/_{√(9−𝑥²)}

arcsin⁢(𝑥/3)+𝐶

∫ 𝑑𝑥/_9+𝑥²

1/3arctan⁢(𝑥/3)+𝐶

∫arcsin(t)dt/_√(1-t²)

1/2⁢(arcsin⁢𝑡)²+𝐶

∫e^t/_{√(1-e^2t)dt}

arcsin(𝑒^𝑡)+𝐶

∫₀^½sin(arctan(t))/_(1+t²)dt

1−2/√5