What is the integral of 4xarcsin(x^2) ?

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

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integral of 4xarcsin(x^2)

\int 4 x arcsin (x^{2}) d x

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

Solution steps

\int 4 x arcsin (x^{2}) d x

Take the constant out:

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

= 4 \cdot \int x arcsin (x^{2}) d x

Apply Integration By Parts

: \frac{1}{2} x^{2} arcsin (x^{2}) - \int \frac{x ^{3}}{1 - x ^{4}} d x

= 4 (\frac{1}{2} x^{2} arcsin (x^{2}) - \int \frac{x ^{3}}{1 - x ^{4}} d x)

\int \frac{x ^{3}}{1 - x ^{4}} d x = - \frac{1}{2} 1 - x^{4}

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

Simplify

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

Add a constant to the solution

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

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