What is the integral of (x^2)/((1-y^2)) ?

The integral of (x^2)/((1-y^2)) is (x^3)/(3(1-y^2))+C

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integral of (x^2)/((1-y^2))

\int \frac{x ^{2}}{( 1 - y ^{2} )}

\frac{x ^{3}}{3 ( 1 - y ^{2} )} + C

Solution steps

\int \frac{x ^{2}}{1 - y ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{1 - y ^{2}} \cdot \int x^{2} d x

Apply the Power Rule

= \frac{1}{1 - y ^{2}} \cdot \frac{x ^{3}}{3}

Simplify

\frac{1}{1 - y ^{2}} \cdot \frac{x ^{3}}{3} : \frac{x ^{3}}{3 ( 1 - y ^{2} )}

= \frac{x ^{3}}{3 ( 1 - y ^{2} )}

Add a constant to the solution

= \frac{x ^{3}}{3 ( 1 - y ^{2} )} + C

What is the integral of (x^2)/((1-y^2)) ?
The integral of (x^2)/((1-y^2)) is (x^3)/(3(1-y^2))+C