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)#