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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int sin (u) + u d u

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

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

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

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

Substitute back

u = arcsin (x)

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{( x ^{2} )}{2 + x})

\frac{1}{x - 4}

x \to 1 + lim (\frac{2}{x} - \frac{1}{ln ( x )})

\frac{\partial}{\partial x} (2 y \cdot arctan (\frac{x}{y}))

6 x + 4

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