What is the integral of 6/(5zsqrt(4z^4-9)) ?

The integral of 6/(5zsqrt(4z^4-9)) is 1/5 arctan((sqrt(4z^4-9))/3)+C

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integral of 6/(5zsqrt(4z^4-9))

\int \frac{6}{5 z 4 z ^{4} - 9} d z

\frac{1}{5} arctan (\frac{4 z ^{4} - 9}{3}) + C

Solution steps

\int \frac{6}{5 z 4 z ^{4} - 9} d z

Take the constant out:

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

= \frac{6}{5} \cdot \int \frac{1}{z 4 z ^{4} - 9} d z

Apply u-substitution

: \int \frac{1}{4 u 4 u - 9} d u

= \frac{6}{5} \cdot \int \frac{1}{4 u 4 u - 9} d u

Take the constant out:

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

= \frac{6}{5} \cdot \frac{1}{4} \cdot \int \frac{1}{u 4 u - 9} d u

Apply u-substitution

: \int \frac{2}{v ^{2} + 9} d v

= \frac{6}{5} \cdot \frac{1}{4} \cdot \int \frac{2}{v ^{2} + 9} d v

Take the constant out:

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

= \frac{6}{5} \cdot \frac{1}{4} \cdot 2 \cdot \int \frac{1}{v ^{2} + 9} d v

Apply Integral Substitution

= \frac{6}{5} \cdot \frac{1}{4} \cdot 2 \cdot \int \frac{1}{3 ( w ^{2} + 1 )} d w

Take the constant out:

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

= \frac{6}{5} \cdot \frac{1}{4} \cdot 2 \cdot \frac{1}{3} \cdot \int \frac{1}{w ^{2} + 1} d w

Use the common integral:

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

= \frac{6}{5} \cdot \frac{1}{4} \cdot 2 \cdot \frac{1}{3} arctan (w)

Substitute back

= \frac{6}{5} \cdot \frac{1}{4} \cdot 2 \cdot \frac{1}{3} arctan (\frac{4 z ^{4} - 9}{3})

Simplify

\frac{6}{5} \cdot \frac{1}{4} \cdot 2 \cdot \frac{1}{3} arctan (\frac{4 z ^{4} - 9}{3}) : \frac{1}{5} arctan (\frac{4 z ^{4} - 9}{3})

= \frac{1}{5} arctan (\frac{4 z ^{4} - 9}{3})

Add a constant to the solution

= \frac{1}{5} arctan (\frac{4 z ^{4} - 9}{3}) + C

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What is the integral of 6/(5zsqrt(4z^4-9)) ?
The integral of 6/(5zsqrt(4z^4-9)) is 1/5 arctan((sqrt(4z^4-9))/3)+C