What is the integral of y/(sqrt(25-16y^4)) ?

The integral of y/(sqrt(25-16y^4)) is 1/8 arcsin(4/5 y^2)+C

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integral of y/(sqrt(25-16y^4))

\int \frac{y}{25 - 16 y ^{4}} d y

\frac{1}{8} arcsin (\frac{4}{5} y^{2}) + C

Solution steps

\int \frac{y}{25 - 16 y ^{4}} d y

Apply u-substitution

: \int \frac{1}{2 25 - 16 u ^{2}} d u

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

Apply Trigonometric Substitution

: \int \frac{1}{4} d v

= \frac{1}{2} \cdot \int \frac{1}{4} d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{4} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot \frac{1}{4} \cdot 1 \cdot v

Substitute back

= \frac{1}{2} \cdot \frac{1}{4} \cdot 1 \cdot arcsin (\frac{4}{5} y^{2})

Simplify

\frac{1}{2} \cdot \frac{1}{4} \cdot 1 \cdot arcsin (\frac{4}{5} y^{2}) : \frac{1}{8} arcsin (\frac{4}{5} y^{2})

= \frac{1}{8} arcsin (\frac{4}{5} y^{2})

Add a constant to the solution

= \frac{1}{8} arcsin (\frac{4}{5} y^{2}) + C

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What is the integral of y/(sqrt(25-16y^4)) ?
The integral of y/(sqrt(25-16y^4)) is 1/8 arcsin(4/5 y^2)+C