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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int u + sin (u) d u

= \int u + sin (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int u d u + \int sin (u) d u

\int u d u = \frac{u ^{2}}{2}

\int sin (u) d u = - cos (u)

= \frac{u ^{2}}{2} - cos (u)

Substitute back

u = arcsin (x)

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{x + 4}{1 - x ^{2}} d x

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

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

\int \frac{x y - 1}{x} d x

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

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