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Math - Mixed Integral Practice Common Techniques include: u substitution (both u and matching du must be present) PIntegration by parts (LIATE priority for u in -[vdu) ?· Trig integration (du to set aside: sinx cos x,sec 2 x sec x tan x csc 2 x csc x cot x) R

Math Jan 21, 2021

Math - Mixed Integral Practice Common Techniques include: u substitution (both u and matching du must be present) PIntegration by parts (LIATE priority for u in -[vdu) ?· Trig integration (du to set aside: sinx cos x,sec 2 x sec x tan x csc 2 x csc x cot x) R. F Partial fraction decomposition Radical integration (substitution with trig functions in squared identities) 16. +??! 17. sin 2r dr

Expert Solution

$\int xln(x+1)dx$

let assume that

u=x+1

$\frac{du}{dx}=\frac{d}{dx}(x+1)$

$\frac{du}{dx}=1+0$

du=dx

so we can sy that,

$\int xln(x+1)dx=\int (u-1)ln(u)du$

let assume that,

u=ln(u)

$u'=\frac{d}{du}ln(u)$

$u'=\frac{1}{u}$

and

v'=u-1

$\int v'=\int (u-1)du$

$v=\frac{u^2}{2}-u$

we know that,

$\int uv'=uv-\int u'v$

we can say that,

$\int xln(x+1)dx=ln(u)\left ( \frac{u^2}{2}-u \right )-\int \frac{1}{u}\left ( \frac{u^2}{2}-u \right )du$

$\int xln(x+1)dx=ln(u)\left ( \frac{u^2}{2}-u \right )-\int\left ( \frac{u}{2}-1 \right )du$

$\int xln(x+1)dx=ln(u)\left ( \frac{u^2}{2}-u \right )-\left ( \frac{u^2}{4}-u \right )+C$

$\int xln(x+1)dx=\left ( \frac{u^2}{2}-u \right )ln(u)-\frac{u^2}{4}+u+C$

substitute back u = x+1,

$\int xln(x+1)dx=\left ( \frac{(x+1)^2}{2}-(x+1) \right )ln(u)-\frac{(x+1)^2}{4}+x+1+C$

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