What is the integral of 6x^2arcsin(2x) ?

The integral of 6x^2arcsin(2x) is 2x^3arcsin(2x)+1/6 (2x^2+1)sqrt(-4x^2+1)+C

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integral of 6x^2arcsin(2x)

\int 6 x^{2} arcsin (2 x) d x

2 x^{3} arcsin (2 x) + \frac{1}{6} (2 x^{2} + 1) - 4 x^{2} + 1 + C

Solution steps

\int 6 x^{2} arcsin (2 x) d x

Take the constant out:

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

= 6 \cdot \int x^{2} arcsin (2 x) d x

Apply u-substitution

= 6 \cdot \int \frac{u ^{2} arcsin ( u )}{8} d u

Take the constant out:

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

= 6 \cdot \frac{1}{8} \cdot \int u^{2} arcsin (u) d u

Apply Integration By Parts

= 6 \cdot \frac{1}{8} (\frac{1}{3} u^{3} arcsin (u) - \int \frac{u ^{3}}{3 1 - u ^{2}} d u)

\int \frac{u ^{3}}{3 1 - u ^{2}} d u = \frac{1}{3} (\frac{1}{3} (1 - u^{2})^{\frac{3}{2}} - 1 - u^{2})

= 6 \cdot \frac{1}{8} (\frac{1}{3} u^{3} arcsin (u) - \frac{1}{3} (\frac{1}{3} (1 - u^{2})^{\frac{3}{2}} - 1 - u^{2}))

Substitute back

u = 2 x

= 6 \cdot \frac{1}{8} (\frac{1}{3} (2 x)^{3} arcsin (2 x) - \frac{1}{3} (\frac{1}{3} (1 - (2 x)^{2})^{\frac{3}{2}} - 1 - (2 x)^{2}))

Simplify

6 \cdot \frac{1}{8} (\frac{1}{3} (2 x)^{3} arcsin (2 x) - \frac{1}{3} (\frac{1}{3} (1 - (2 x)^{2})^{\frac{3}{2}} - 1 - (2 x)^{2})) : 2 x^{3} arcsin (2 x) + \frac{1}{6} (2 x^{2} + 1) - 4 x^{2} + 1

= 2 x^{3} arcsin (2 x) + \frac{1}{6} (2 x^{2} + 1) - 4 x^{2} + 1

Add a constant to the solution

= 2 x^{3} arcsin (2 x) + \frac{1}{6} (2 x^{2} + 1) - 4 x^{2} + 1 + C

What is the integral of 6x^2arcsin(2x) ?
The integral of 6x^2arcsin(2x) is 2x^3arcsin(2x)+1/6 (2x^2+1)sqrt(-4x^2+1)+C