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

The integral of ((x^2))/(1-x^2) is (ln|x+1|)/2-(ln|x-1|)/2-x+C

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

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

\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2} - x + C

Solution steps

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

\frac{x ^{2}}{1 - x ^{2}} = \frac{1}{1 - x ^{2}} - 1

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{1 - x ^{2}} d x = \frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}

\int 1 d x = x

= \frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2} - x

Add a constant to the solution

= \frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2} - x + C

\int \frac{sin ( x ) - cos ( x )}{cos ( x )} d x

\int sec (x) (5 sec (x) + 7 tan (x)) d x

\int x^{3} + 3 x d x

\int \frac{arccos ( x )}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

\int x^{2} - 3 x d x

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