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

The integral is here:

Divide the fraction in half, followed by the integrals:

Utilize the power rule for integrals to integrate:

To further simplify:

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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.

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