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

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

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integral of (xarcsin(x^2))/(sqrt(1-x^4))

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

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

Solution steps

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

Apply u-substitution

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

Apply u-substitution

= \frac{1}{2} \cdot \int v d v

Apply the Power Rule

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

Substitute back

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

Simplify

\frac{1}{2} \cdot \frac{arcsin ^{2} ( x ^{2} )}{2} : \frac{1}{4} arcsin^{2} (x^{2})

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

Add a constant to the solution

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

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