What is the antiderivative of #x^3/(1+x^2)#?

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Answer 1 · 2016-07-07T16:51:03

#1/2{x^2+1-ln(x^2+1)}+C.#

OR

#1/2{x^2-ln(x^2+1)}+C_1,# where #C_1=C+1/2.#

Let #I=intx^3/(1+x^2)dx#

We take substn. #x^2+1=t#, so that, #2xdx=dt.#

Also, #x^2+1=t rArr x^2=t-1#

Now, #I=intx^3/(1+x^2)dx=1/2int(x^2*2x)/(1+x^2)dx=1/2int(t-1)/tdt#
#=1/2int{t/t-1/t}dt=1/2int{1-1/t}dt=1/2{t-lnt}=1/2{x^2+1-ln(x^2+1)}+C.#

#I# is also#=1/2x^2+1/2-1/2ln(x^2+1)+C#
#=1/2{x^2-ln(x^2+1)}+C_1,# where #C_1=C+1/2#

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