Answer to Question #100022 in Calculus for Jayson

Question

Integration by parts

1. Integral of 2x tan^-1 (x) dx

2. Integral of x^2 (x+4)^5 dx

Answer 1

$1. \int2x arctan(x)dx= 2 \int x arctan(x) dx = \begin{vmatrix} u = arctan(x) \\ dv = xdx\\ du = \frac 1{x^2+1}dx\\ v = \frac {x^2}2 \end{vmatrix} =$

$=x^2arctan(x) - \int \frac {x^2}{x^2+1}dx =x^2arctan(x) - \int (1 - \frac {1}{x^2+1})dx =\\=x^2arctan(x) - \int1dx+\int\frac{1}{x^2+1} dx =\\$

$= x^2arctan(x)+arctan(x)-x +C$

$2. \int x^2(x+4)^5dx = \int (x^7+20x^6+160x^5+640x^4+1280x^3+1024x^2)dx=$

$=\int x^7dx 20\int x^6 dx+160\int x^5dx +640 \int x^4 dx+1280 \int x^3dx+1024\int x^2dx=\\$$\\= \frac {x^8}8 +\frac {20x^7}7 + \frac {80x^6}3 + 128x^5 + 320x^4 + \frac{1024x^3}3 +C$