What is the integral of (4x)/((3x^2-1)^5) ?

The integral of (4x)/((3x^2-1)^5) is -1/(6(3x^2-1)^4)+C

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integral of (4x)/((3x^2-1)^5)

\int \frac{4 x}{( 3 x ^{2} - 1 ) ^{5}} d x

- \frac{1}{6 ( 3 x ^{2} - 1 ) ^{4}} + C

Solution steps

\int \frac{4 x}{( 3 x ^{2} - 1 ) ^{5}} d x

Take the constant out:

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

= 4 \cdot \int \frac{x}{( 3 x ^{2} - 1 ) ^{5}} d x

Apply u-substitution

: \int \frac{1}{2 ( 3 u - 1 ) ^{5}} d u

= 4 \cdot \int \frac{1}{2 ( 3 u - 1 ) ^{5}} d u

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \int \frac{1}{( 3 u - 1 ) ^{5}} d u

Apply u-substitution

: \int \frac{1}{3 v ^{5}} d v

= 4 \cdot \frac{1}{2} \cdot \int \frac{1}{3 v ^{5}} d v

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \int \frac{1}{v ^{5}} d v

Apply the Power Rule

: - \frac{1}{4 v ^{4}}

= 4 \cdot \frac{1}{2} \cdot \frac{1}{3} (- \frac{1}{4 v ^{4}})

Substitute back

= 4 \cdot \frac{1}{2} \cdot \frac{1}{3} (- \frac{1}{4 ( 3 x ^{2} - 1 ) ^{4}})

Simplify

4 \cdot \frac{1}{2} \cdot \frac{1}{3} (- \frac{1}{4 ( 3 x ^{2} - 1 ) ^{4}}) : - \frac{1}{6 ( 3 x ^{2} - 1 ) ^{4}}

= - \frac{1}{6 ( 3 x ^{2} - 1 ) ^{4}}

Add a constant to the solution

= - \frac{1}{6 ( 3 x ^{2} - 1 ) ^{4}} + C

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What is the integral of (4x)/((3x^2-1)^5) ?
The integral of (4x)/((3x^2-1)^5) is -1/(6(3x^2-1)^4)+C