What is the integral of+y/(sqrt(16-9y^4)) ?

The integral of+y/(sqrt(16-9y^4)) is 1/6 arcsin(3/4 y^2)+C

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integral of+y/(sqrt(16-9y^4))

\int + \frac{y}{16 - 9 y ^{4}} d y

\frac{1}{6} arcsin (\frac{3}{4} y^{2}) + C

Solution steps

\int \frac{y}{16 - 9 y ^{4}} d y

Apply u-substitution

: \int \frac{1}{2 16 - 9 u ^{2}} d u

= \int \frac{1}{2 16 - 9 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}{16 - 9 u ^{2}} d u

Apply Trigonometric Substitution

: \int \frac{1}{3} d v

= \frac{1}{2} \cdot \int \frac{1}{3} 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}{3} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot \frac{1}{3} \cdot 1 \cdot v

Substitute back

= \frac{1}{2} \cdot \frac{1}{3} \cdot 1 \cdot arcsin (\frac{3}{4} y^{2})

Simplify

\frac{1}{2} \cdot \frac{1}{3} \cdot 1 \cdot arcsin (\frac{3}{4} y^{2}) : \frac{1}{6} arcsin (\frac{3}{4} y^{2})

= \frac{1}{6} arcsin (\frac{3}{4} y^{2})

Add a constant to the solution

= \frac{1}{6} arcsin (\frac{3}{4} y^{2}) + C

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What is the integral of+y/(sqrt(16-9y^4)) ?
The integral of+y/(sqrt(16-9y^4)) is 1/6 arcsin(3/4 y^2)+C