What is the integral of (x^7)/(sqrt(1-25x^{16))} ?

The integral of (x^7)/(sqrt(1-25x^{16))} is 1/40 arcsin(5x^8)+C

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integral of (x^7)/(sqrt(1-25x^{16))}

\int \frac{x ^{7}}{1 - 25 x ^{16}} d x

\frac{1}{40} arcsin (5 x^{8}) + C

Solution steps

\int \frac{x ^{7}}{1 - 25 x ^{16}} d x

Apply u-substitution

: \int \frac{1}{8 1 - 25 u ^{2}} d u

= \int \frac{1}{8 1 - 25 u ^{2}} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{1}{1 - 25 u ^{2}} d u

Apply Trigonometric Substitution

: \int \frac{1}{5} d v

= \frac{1}{8} \cdot \int \frac{1}{5} d v

Take the constant out:

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

= \frac{1}{8} \cdot \frac{1}{5} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{8} \cdot \frac{1}{5} \cdot 1 \cdot v

Substitute back

= \frac{1}{8} \cdot \frac{1}{5} \cdot 1 \cdot arcsin (5 x^{8})

Simplify

\frac{1}{8} \cdot \frac{1}{5} \cdot 1 \cdot arcsin (5 x^{8}) : \frac{1}{40} arcsin (5 x^{8})

= \frac{1}{40} arcsin (5 x^{8})

Add a constant to the solution

= \frac{1}{40} arcsin (5 x^{8}) + C

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What is the integral of (x^7)/(sqrt(1-25x^{16))} ?
The integral of (x^7)/(sqrt(1-25x^{16))} is 1/40 arcsin(5x^8)+C