How do you integrate #int x^3*dx/(x^2-1)^(2/3)#?

Question

Isaiah Allen · Accepted Answer

#intx^3/(x^2-1)^(2/3)dx=(3(x^2-1)^(1/3)(x^2+3))/8+C#

The integral is here:

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

Using substitution, let #u=x^2-1#, so #du=2xdx# and #x^2=u+1#. Thus:

#I=int(x^2(x))/(x^2-1)^(2/3)dx=1/2int(x^2(2x))/(x^2-1)^(2/3)dx=1/2int(u+1)/u^(2/3)du#

Divide the fraction in half, followed by the integrals:

#I=1/2int(u^(1/3)+u^(-2/3))du=1/2intu^(1/3)du+1/2intu^(-2/3)du#

Utilize the power rule for integrals to integrate:

#I=1/2(u^(4/3)/(4/3))+1/2(u^(1/3)/(1/3))=3/8u^(4/3)+3/2u^(1/3)=(3u^(4/3)+12u^(1/3))/8#

To further simplify:

#I=(3u^(1/3)(u+4))/8=(3(x^2-1)^(1/3)((x^2-1)+4))/8=(3(x^2-1)^(1/3)(x^2+3))/8+C#

Aurora Alvarado · Answer

undefined To integrate $ \int \frac{x^3}{(x^2-1)^{2/3}} \, dx $, you can use the substitution method. Let $ u = x^2 - 1 $. Then, $ du = 2x \, dx $. Rewrite the integral in terms of $ u $ and $ du $. This yields:

$$ \int \frac{x^3}{(x^2-1)^{2/3}} \, dx = \frac{1}{2} \int \frac{1}{u^{2/3}} \, du $$

Now, integrate $ \frac{1}{u^{2/3}} $ with respect to $ u $. This gives:

$$ \frac{1}{2} \int \frac{1}{u^{2/3}} \, du = \frac{1}{2} \cdot \frac{u^{1/3}}{1/3} + C $$

Replace $ u $ with $ x^2 - 1 $ to obtain the final result:

$$ \frac{1}{2} \cdot \frac{(x^2 - 1)^{1/3}}{1/3} + C = \frac{3(x^2 - 1)^{1/3}}{2} + C $$

Where $ C $ is the constant of integration.