How do you integrate $\int x ln (1 + x)$ $int xln(1+x)$ using integration by parts?

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Answer 1 · 2016-10-22T09:12:34

$I = \frac{1}{2} x^{2} ln (1 + x) - \frac{1}{4} x^{2} + \frac{1}{2} x - \frac{1}{2} ln (1 + x) + C$ $I=1/2x^2ln(1+x)-1/4x^2+1/2x-1/2ln(1+x)+C$

$\int u \frac{d v}{d x} d x = u v - \int v \frac{d u}{d x} d x$ $intu(dv)/dxdx=uv-intv(du)/dxdx$

let $u = ln (1 + x) \Rightarrow \frac{d u}{d x} = \frac{1}{1 + x}$ $u=ln(1+x)=>(du)/dx=1/(1+x)$

$\frac{d v}{d x} = x \Rightarrow v = \frac{1}{2} x^{2}$ $(dv)/dx=x=>v=1/2x^2$

$I = \frac{1}{2} x^{2} ln (1 + x) - \frac{1}{2} \int (\frac{x^{2}}{1 + x}) d x$ $I=1/2x^2ln(1+x)-1/2int(x^2/(1+x))dx$

$I = \frac{1}{2} x^{2} ln (1 + x) - \frac{1}{2} \int (x - \frac{x}{1 + x}) d x$ $I=1/2x^2ln(1+x)-1/2int(x-x/(1 +x))dx$

$I = \frac{1}{2} x^{2} ln (1 + x) - \frac{1}{2} \int (x - 1 + \frac{1}{x + 1}) d x$ $I=1/2x^2ln(1+x)-1/2int(x-1+1/(x+1))dx$

$I = \frac{1}{2} x^{2} ln (1 + x) - \frac{1}{4} x^{2} + \frac{1}{2} x - \frac{1}{2} ln (1 + x) + C$ $I=1/2x^2ln(1+x)-1/4x^2+1/2x-1/2ln(1+x)+C$

How do you integrate int xln(1+x)∫xln(1+x)int xln(1+x) using integration by parts?