The problem asks us to evaluate several integrals and then to perform a partial fraction decomposition and evaluate the resulting integral. Specifically, we need to evaluate: i. $\int \frac{x^3}{x+1} dx$ ii. $\int (3x+5)^{-2} dx$ iii. $\int \cos(4x) e^{\sin(4x)} dx$ iv. $\int \frac{1}{x^2 - 6x + 16} dx$ v. $\int \frac{\sin^{-1}(x)}{\sqrt{1-x^2}} dx$ vi. $\int (x-1) \cot(x^2-2x-1) dx$ And for the second problem: Given $\frac{3x+7}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$, find $A$ and $B$, and hence evaluate $\int \frac{3x+7}{(x-1)(x+2)} dx$.

1. Problem Description

The problem asks us to evaluate several integrals and then to perform a partial fraction decomposition and evaluate the resulting integral. Specifically, we need to evaluate:

i. $\int \frac{x^3}{x+1} dx$

ii. $\int (3x+5)^{-2} dx$

iii. $\int \cos(4x) e^{\sin(4x)} dx$

iv. $\int \frac{1}{x^2 - 6x + 16} dx$

v. $\int \frac{\sin^{-1}(x)}{\sqrt{1-x^2}} dx$

vi. $\int (x-1) \cot(x^2-2x-1) dx$

And for the second problem:

Given $\frac{3x+7}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$, find $A$ and $B$, and hence evaluate $\int \frac{3x+7}{(x-1)(x+2)} dx$.

2. Solution Steps

i. $\int \frac{x^3}{x+1} dx$:

We can perform polynomial long division to get $x^3 = (x+1)(x^2 - x + 1) - 1$.

So $\frac{x^3}{x+1} = x^2 - x + 1 - \frac{1}{x+1}$.

$\int \frac{x^3}{x+1} dx = \int (x^2 - x + 1 - \frac{1}{x+1}) dx = \frac{x^3}{3} - \frac{x^2}{2} + x - \ln|x+1| + C$.

ii. $\int (3x+5)^{-2} dx$:

Let $u = 3x+5$. Then $du = 3 dx$, so $dx = \frac{1}{3} du$.

$\int (3x+5)^{-2} dx = \int u^{-2} \frac{1}{3} du = \frac{1}{3} \int u^{-2} du = \frac{1}{3} \frac{u^{-1}}{-1} + C = -\frac{1}{3u} + C = -\frac{1}{3(3x+5)} + C$.

iii. $\int \cos(4x) e^{\sin(4x)} dx$:

Let $u = \sin(4x)$. Then $du = 4\cos(4x) dx$, so $\cos(4x) dx = \frac{1}{4} du$.

$\int \cos(4x) e^{\sin(4x)} dx = \int e^u \frac{1}{4} du = \frac{1}{4} \int e^u du = \frac{1}{4} e^u + C = \frac{1}{4} e^{\sin(4x)} + C$.

iv. $\int \frac{1}{x^2 - 6x + 16} dx$:

Complete the square: $x^2 - 6x + 16 = (x^2 - 6x + 9) + 7 = (x-3)^2 + 7$.

$\int \frac{1}{x^2 - 6x + 16} dx = \int \frac{1}{(x-3)^2 + 7} dx$.

Let $u = x-3$, so $du = dx$. Then $\int \frac{1}{u^2 + 7} du = \frac{1}{\sqrt{7}} \arctan(\frac{u}{\sqrt{7}}) + C = \frac{1}{\sqrt{7}} \arctan(\frac{x-3}{\sqrt{7}}) + C$.

v. $\int \frac{\sin^{-1}(x)}{\sqrt{1-x^2}} dx$:

Let $u = \sin^{-1}(x)$. Then $du = \frac{1}{\sqrt{1-x^2}} dx$.

$\int \frac{\sin^{-1}(x)}{\sqrt{1-x^2}} dx = \int u du = \frac{u^2}{2} + C = \frac{(\sin^{-1}(x))^2}{2} + C$.

vi. $\int (x-1) \cot(x^2-2x-1) dx$:

Let $u = x^2-2x-1$. Then $du = (2x-2) dx = 2(x-1) dx$, so $(x-1) dx = \frac{1}{2} du$.

$\int (x-1) \cot(x^2-2x-1) dx = \int \cot(u) \frac{1}{2} du = \frac{1}{2} \int \cot(u) du = \frac{1}{2} \int \frac{\cos(u)}{\sin(u)} du$.

Let $v = \sin(u)$, so $dv = \cos(u) du$.

$\frac{1}{2} \int \frac{1}{v} dv = \frac{1}{2} \ln|v| + C = \frac{1}{2} \ln|\sin(u)| + C = \frac{1}{2} \ln|\sin(x^2-2x-1)| + C$.

For the second problem:

$\frac{3x+7}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$.

$3x+7 = A(x+2) + B(x-1)$.

If $x = 1$, $3(1) + 7 = A(1+2) + B(1-1)$, so $10 = 3A$, and $A = \frac{10}{3}$.

If $x = -2$, $3(-2) + 7 = A(-2+2) + B(-2-1)$, so $1 = -3B$, and $B = -\frac{1}{3}$.

$\int \frac{3x+7}{(x-1)(x+2)} dx = \int (\frac{10/3}{x-1} - \frac{1/3}{x+2}) dx = \frac{10}{3} \int \frac{1}{x-1} dx - \frac{1}{3} \int \frac{1}{x+2} dx = \frac{10}{3} \ln|x-1| - \frac{1}{3} \ln|x+2| + C$.

3. Final Answer

i. $\frac{x^3}{3} - \frac{x^2}{2} + x - \ln|x+1| + C$

ii. $-\frac{1}{3(3x+5)} + C$

iii. $\frac{1}{4} e^{\sin(4x)} + C$

iv. $\frac{1}{\sqrt{7}} \arctan(\frac{x-3}{\sqrt{7}}) + C$

v. $\frac{(\sin^{-1}(x))^2}{2} + C$

vi. $\frac{1}{2} \ln|\sin(x^2-2x-1)| + C$

$A = \frac{10}{3}$, $B = -\frac{1}{3}$

$\int \frac{3x+7}{(x-1)(x+2)} dx = \frac{10}{3} \ln|x-1| - \frac{1}{3} \ln|x+2| + C$