Question #100022
Integration by parts

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

2. Integral of x^2 (x+4)^5 dx
1
Expert's answer
2019-12-11T11:05:14-0500

1.2xarctan(x)dx=2xarctan(x)dx=u=arctan(x)dv=xdxdu=1x2+1dxv=x22=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} =

=x2arctan(x)x2x2+1dx=x2arctan(x)(11x2+1)dx==x2arctan(x)1dx+1x2+1dx==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 =\\

=x2arctan(x)+arctan(x)x+C= x^2arctan(x)+arctan(x)-x +C

2.x2(x+4)5dx=(x7+20x6+160x5+640x4+1280x3+1024x2)dx=2. \int x^2(x+4)^5dx = \int (x^7+20x^6+160x^5+640x^4+1280x^3+1024x^2)dx=

=x7dx20x6dx+160x5dx+640x4dx+1280x3dx+1024x2dx==\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=\\=x88+20x77+80x63+128x5+320x4+1024x33+C\\= \frac {x^8}8 +\frac {20x^7}7 + \frac {80x^6}3 + 128x^5 + 320x^4 + \frac{1024x^3}3 +C


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