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

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

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

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

ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - x + C

Solution steps

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

Simplify

1 + (\frac{2 x}{1 - x ^{2}})^{2} : \frac{x ^{2} + 1}{1 - x ^{2}}

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

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

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

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

\int 1 d x = x

= ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - x

Add a constant to the solution

= ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - x + C

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