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

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

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

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

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

Solution steps

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

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

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

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

\int 1 d x = x

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

Add a constant to the solution

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

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