What is the integral of (x^4)/(sqrt(1-4x^{10))} ?

The integral of (x^4)/(sqrt(1-4x^{10))} is 1/10 arcsin(2x^5)+C

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integral of (x^4)/(sqrt(1-4x^{10))}

\int \frac{x ^{4}}{1 - 4 x ^{10}} d x

\frac{1}{10} arcsin (2 x^{5}) + C

Solution steps

\int \frac{x ^{4}}{1 - 4 x ^{10}} d x

Apply u-substitution

: \int \frac{1}{5 1 - 4 u ^{2}} d u

= \int \frac{1}{5 1 - 4 u ^{2}} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int \frac{1}{1 - 4 u ^{2}} d u

Apply Trigonometric Substitution

: \int \frac{1}{2} d v

= \frac{1}{5} \cdot \int \frac{1}{2} d v

Take the constant out:

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

= \frac{1}{5} \cdot \frac{1}{2} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{5} \cdot \frac{1}{2} \cdot 1 \cdot v

Substitute back

= \frac{1}{5} \cdot \frac{1}{2} \cdot 1 \cdot arcsin (2 x^{5})

Simplify

\frac{1}{5} \cdot \frac{1}{2} \cdot 1 \cdot arcsin (2 x^{5}) : \frac{1}{10} arcsin (2 x^{5})

= \frac{1}{10} arcsin (2 x^{5})

Add a constant to the solution

= \frac{1}{10} arcsin (2 x^{5}) + C

\int x (5 x^{2} - 1)^{3} d x

\int \frac{2 x ^{2} + x - 1}{1 + x ^{2}} d x

\int cos (\frac{1}{10} x) d x

\int x^{2} \frac{2}{1 + x ^{3}} d x

\int \frac{1}{0} d x

What is the integral of (x^4)/(sqrt(1-4x^{10))} ?
The integral of (x^4)/(sqrt(1-4x^{10))} is 1/10 arcsin(2x^5)+C