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

The integral of (x^2+2)/(sqrt(1-x^2)) is 5/2 arcsin(x)-1/4 sin(2arcsin(x))+C

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

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

\frac{5}{2} arcsin (x) - \frac{1}{4} sin (2 arcsin (x)) + C

Solution steps

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

Expand

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

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

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

Simplify

\frac{1}{2} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x))) + 2 arcsin (x) : \frac{5}{2} arcsin (x) - \frac{1}{4} sin (2 arcsin (x))

= \frac{5}{2} arcsin (x) - \frac{1}{4} sin (2 arcsin (x))

Add a constant to the solution

= \frac{5}{2} arcsin (x) - \frac{1}{4} sin (2 arcsin (x)) + C

What is the integral of (x^2+2)/(sqrt(1-x^2)) ?
The integral of (x^2+2)/(sqrt(1-x^2)) is 5/2 arcsin(x)-1/4 sin(2arcsin(x))+C