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

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

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

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

\frac{1}{2} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x))) - 1 - x^{2} + C

Solution steps

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

Expand

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

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

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

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

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

Add a constant to the solution

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

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