What is the integral of (arcsin(5x))/(sqrt(1-25x^2)) ?

The integral of (arcsin(5x))/(sqrt(1-25x^2)) is 1/10 arcsin^2(5x)+C

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integral of (arcsin(5x))/(sqrt(1-25x^2))

\int \frac{arcsin ( 5 x )}{1 - 25 x ^{2}} d x

\frac{1}{10} arcsin^{2} (5 x) + C

Solution steps

\int \frac{arcsin ( 5 x )}{1 - 25 x ^{2}} d x

Apply u-substitution

: \int \frac{arcsin ( u )}{5 1 - u ^{2}} d u

= \int \frac{arcsin ( u )}{5 1 - u ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{5} \cdot \int \frac{arcsin ( u )}{1 - u ^{2}} d u

Apply u-substitution

: \int v d v

= \frac{1}{5} \cdot \int v d v

Apply the Power Rule

: \frac{v ^{2}}{2}

= \frac{1}{5} \cdot \frac{v ^{2}}{2}

Substitute back

= \frac{1}{5} \cdot \frac{arcsin ^{2} ( 5 x )}{2}

Simplify

\frac{1}{5} \cdot \frac{arcsin ^{2} ( 5 x )}{2} : \frac{1}{10} arcsin^{2} (5 x)

= \frac{1}{10} arcsin^{2} (5 x)

Add a constant to the solution

= \frac{1}{10} arcsin^{2} (5 x) + C

t^{2} y^{^{''}} + 11 t y^{^{'}} + 25 y = 0

\frac{\partial}{\partial x} (e^{- 5 x^{2} - 2 y^{2}})

\int \frac{t ^{3} + 2}{t ^{4} + 8 t - 3} d t

x \to \infty - lim (e^{x})

\int_{0}^{x} e^{x + y} d y

What is the integral of (arcsin(5x))/(sqrt(1-25x^2)) ?
The integral of (arcsin(5x))/(sqrt(1-25x^2)) is 1/10 arcsin^2(5x)+C