How do you find the integral of $int x^3/(x^2+1)dx$ ?

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Answer 1 · 2016-12-03T13:03:40

$intx^3/(1+x^2)dx = 1/2x^2-lnsqrt(1+x^2)$

Substitute:

$x=tant$

$dx = (dt)/cos^2t$

$intx^3/(1+x^2)dx = int tan^3t/(1+tan^2t) dt/cos^2t$

Use the trigonometric identity:

$1+tan^2 t = 1/cos^2t$

$intx^3/(1+x^2)dx = int tan^3t/( 1/cos^2t) dt/cos^2t = int tan^3tdt$

Using the same identity again:

$int tan^3tdt = int tant tan^2t dt = int tant (1/cos^2t -1)dt =$

$= int tant d(tant) - int tant dt =1/2 tan^2 t - int sint /cost dt =$

$= 1/2tan^2t + int (d(cost))/cost = 1/2tan^2t + ln |cos t|$

To substitute back $x$ , we have that:

$tant = x$

$cost= +-sqrt(1 /(1+tan^2t))= +-1/sqrt(1+x^2)$

So:

$intx^3/(1+x^2)dx = 1/2x^2-lnsqrt(1+x^2)$

How do you find the integral of int x^3/(x^2+1)dxint x^3/(x^2+1)dx?