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

Answer 1

Evelyn Agard

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

#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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Answer 2

Adam Adkins

The antiderivative of (\frac{x^3}{1+x^2}) can be found using the substitution method. Let (u = 1 + x^2), then (du = 2x , dx). Rewriting the integral in terms of (u) gives (\int \frac{x^3}{1+x^2} , dx = \frac{1}{2} \int \frac{u-1}{u} , du). Simplifying and integrating gives (\frac{1}{2} \int \left(1 - \frac{1}{u}\right) , du = \frac{1}{2} (u - \ln|u|) + C = \frac{1}{2}(1 + x^2 - \ln(1 + x^2)) + C), where (C) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.