How do you use substitution to integrate #(4x^3)/( (x^2+1)^2)#?

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Answer 1 · 2018-03-05T22:44:15

I tried this:

Have a look:
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Answer 2 · 2018-03-05T22:53:56

#int (4x^3)/(x^2+1)^2dx = 2ln (1+x^2) +2/(1+x^2)+C#

Substitute #x = tant#, #dx= sec^2t#

#int (4x^3)/(x^2+1)^2dx = int (4tan^3tsec^2t)/(tan^2t+1)^2dt#

Use now the trigonometric identity:

#1+tan^2t = sec^2t#

#int (4x^3)/(x^2+1)^2dx = int (4tan^3tsec^2t)/sec^4tdt#

#int (4x^3)/(x^2+1)^2dx = int (4tan^3t)/sec^2tdt#

#int (4x^3)/(x^2+1)^2dx = int (4sin^3tcos^2)/cos^3tdt#

#int (4x^3)/(x^2+1)^2dx = int (4sin^3t)/costdt#

#int (4x^3)/(x^2+1)^2dx = 4int ((1-cos^2t)sint)/costdt#

#int (4x^3)/(x^2+1)^2dx = -4int (1/cost-cost) d(cost)#

#int (4x^3)/(x^2+1)^2dx = -4ln abs (cost) +2 cos^2t + C#

using the properties of logarithms:

#int (4x^3)/(x^2+1)^2dx = 2ln (1/ cos^2t) +2 cos^2t + C#

To undo the substitution note that:

#cos^2t = 1/sec^2t = 1/(1+tan^2t) = 1/(1+x^2)#

#int (4x^3)/(x^2+1)^2dx = 2ln (1+x^2) +2/(1+x^2)+C#