What is the integral of arcsin(4x) ?

The integral of arcsin(4x) is xarcsin(4x)+1/4 sqrt(1-16x^2)+C

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integral of arcsin(4x)

\int arcsin (4 x) d x

x arcsin (4 x) + \frac{1}{4} 1 - 16 x^{2} + C

Solution steps

\int arcsin (4 x) d x

Apply Integration By Parts

: x arcsin (4 x) - \int \frac{4 x}{1 - 16 x ^{2}} d x

= x arcsin (4 x) - \int \frac{4 x}{1 - 16 x ^{2}} d x

\int \frac{4 x}{1 - 16 x ^{2}} d x = - \frac{1}{4} 1 - 16 x^{2}

= x arcsin (4 x) - (- \frac{1}{4} 1 - 16 x^{2})

Simplify

= x arcsin (4 x) + \frac{1}{4} 1 - 16 x^{2}

Add a constant to the solution

= x arcsin (4 x) + \frac{1}{4} 1 - 16 x^{2} + C

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

x \to 2 + lim (\frac{- 3}{x ^{2} - 4})

y = 2 x + 15, y = x^{2}

f (t) = t + 1

n = 1 \sum \infty \frac{7}{4 n ^{2} - 1}

What is the integral of arcsin(4x) ?
The integral of arcsin(4x) is xarcsin(4x)+1/4 sqrt(1-16x^2)+C