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 5 tan^{3} (x) d x

x \to 0 lim (\frac{2}{( x - 5 ) ^{6}})

- 3 y^{^{'}} + \frac{2 y}{x} = \frac{y ^{4}}{x ^{4}}

y^{^{''}} - 3 y^{^{'}} - 4 y = 12

\frac{\partial}{\partial y} (5 x^{2} + y^{2})

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