What is the integral of (16x^2)/(sqrt(25-36x^6)) ?

The integral of (16x^2)/(sqrt(25-36x^6)) is 8/9 arcsin(6/5 x^3)+C

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integral of (16x^2)/(sqrt(25-36x^6))

\int \frac{16 x ^{2}}{25 - 36 x ^{6}} d x

\frac{8}{9} arcsin (\frac{6}{5} x^{3}) + C

Solution steps

\int \frac{16 x ^{2}}{25 - 36 x ^{6}} d x

Take the constant out:

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

= 16 \cdot \int \frac{x ^{2}}{25 - 36 x ^{6}} d x

Apply u-substitution

: \int \frac{1}{3 25 - 36 u ^{2}} d u

= 16 \cdot \int \frac{1}{3 25 - 36 u ^{2}} d u

Take the constant out:

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

= 16 \cdot \frac{1}{3} \cdot \int \frac{1}{25 - 36 u ^{2}} d u

Apply Trigonometric Substitution

: \int \frac{1}{6} d v

= 16 \cdot \frac{1}{3} \cdot \int \frac{1}{6} d v

Take the constant out:

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

= 16 \cdot \frac{1}{3} \cdot \frac{1}{6} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 16 \cdot \frac{1}{3} \cdot \frac{1}{6} \cdot 1 \cdot v

Substitute back

= 16 \cdot \frac{1}{3} \cdot \frac{1}{6} \cdot 1 \cdot arcsin (\frac{6}{5} x^{3})

Simplify

16 \cdot \frac{1}{3} \cdot \frac{1}{6} \cdot 1 \cdot arcsin (\frac{6}{5} x^{3}) : \frac{8}{9} arcsin (\frac{6}{5} x^{3})

= \frac{8}{9} arcsin (\frac{6}{5} x^{3})

Add a constant to the solution

= \frac{8}{9} arcsin (\frac{6}{5} x^{3}) + C

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What is the integral of (16x^2)/(sqrt(25-36x^6)) ?
The integral of (16x^2)/(sqrt(25-36x^6)) is 8/9 arcsin(6/5 x^3)+C