What is the integral of xarcsin(4x) ?

The integral of xarcsin(4x) is 1/16 (8x^2arcsin(4x)-1/4 (arcsin(4x)-1/2 sin(2arcsin(4x))))+C

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integral of xarcsin(4x)

\int x arcsin (4 x) d x

\frac{1}{16} (8 x^{2} arcsin (4 x) - \frac{1}{4} (arcsin (4 x) - \frac{1}{2} sin (2 arcsin (4 x)))) + C

Solution steps

\int x arcsin (4 x) d x

Apply u-substitution

: \int \frac{u arcsin ( u )}{16} d u

= \int \frac{u arcsin ( u )}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int u arcsin (u) d u

Apply Integration By Parts

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

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

\int \frac{u ^{2}}{2 1 - u ^{2}} d u = \frac{1}{4} (arcsin (u) - \frac{1}{2} sin (2 arcsin (u)))

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

Substitute back

u = 4 x

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

\frac{1}{2} (4 x)^{2} arcsin (4 x) = 8 x^{2} arcsin (4 x)

= \frac{1}{16} (8 x^{2} arcsin (4 x) - \frac{1}{4} (arcsin (4 x) - \frac{1}{2} sin (2 arcsin (4 x))))

Add a constant to the solution

= \frac{1}{16} (8 x^{2} arcsin (4 x) - \frac{1}{4} (arcsin (4 x) - \frac{1}{2} sin (2 arcsin (4 x)))) + C

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What is the integral of xarcsin(4x) ?
The integral of xarcsin(4x) is 1/16 (8x^2arcsin(4x)-1/4 (arcsin(4x)-1/2 sin(2arcsin(4x))))+C