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Answer 1 · 2017-03-19T16:51:06

$xln(x^2+1)-2x+2arc tanx+C.$

Let $I=intln(x^2+1)dx=int{(ln(x^2+1)(1)}dx.$

We use the following Rule of Integration by Parts (IBP) :

$(IBP) : intuvdx=uintvdx-int{((du)/dx)(intvdx)}dx.$

We take $: u=ln(x^2+1) :. (du)/dx=1/(x^2+1)d/dx(x^2+1), i.e.,$

$(du)/dx=(2x)/(x^2+1)." Also, "v=1 rArr intvdx=x.$

Hence, $I=xln(x^2+1)-int{((2x)/(x^2+1))(x)}dx.$

$=xln(x^2+1)-2intx^2/(x^2+1)dx$

$=xln(x^2+1)-2int{(x^2+1)-1}/(x^2+1)dx.$

$=xln(x^2+1)-2int{(x^2+1)/(x^2+1)-1/(x^2+1)}dx.$

$=xln(x^2+1)-2int1dx+2int1/(x^2+1)dx.$

$:. I=xln(x^2+1)-2x+2arc tanx+C.$

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