What is the integral of 3/(ysqrt(y^4-9)) ?

The integral of 3/(ysqrt(y^4-9)) is 1/2 arctan(1/3 sqrt(y^4-9))+C

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integral of 3/(ysqrt(y^4-9))

\int \frac{3}{y y ^{4} - 9} d y

\frac{1}{2} arctan (\frac{1}{3} y^{4} - 9) + C

Solution steps

\int \frac{3}{y y ^{4} - 9} d y

Take the constant out:

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

= 3 \cdot \int \frac{1}{y y ^{4} - 9} d y

Apply u-substitution

: \int \frac{1}{2 ( u ^{2} + 9 )} d u

= 3 \cdot \int \frac{1}{2 ( u ^{2} + 9 )} d u

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{2} + 9} d u

Apply Integral Substitution

= 3 \cdot \frac{1}{2} \cdot \int \frac{1}{3 ( v ^{2} + 1 )} d v

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= 3 \cdot \frac{1}{2} \cdot \frac{1}{3} arctan (v)

Substitute back

= 3 \cdot \frac{1}{2} \cdot \frac{1}{3} arctan (\frac{y ^{4} - 9}{3})

Simplify

3 \cdot \frac{1}{2} \cdot \frac{1}{3} arctan (\frac{y ^{4} - 9}{3}) : \frac{1}{2} arctan (\frac{1}{3} y^{4} - 9)

= \frac{1}{2} arctan (\frac{1}{3} y^{4} - 9)

Add a constant to the solution

= \frac{1}{2} arctan (\frac{1}{3} y^{4} - 9) + C

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