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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Factor

3 x^{2} - 1 : - (- 3 x^{2} + 1)

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

(- (- 3 x^{2} + 1))^{4} = (- 3 x^{2} + 1)^{4}

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

Apply u-substitution

: \int - \frac{1}{6 u ^{4}} d u

= 3 \cdot \int - \frac{1}{6 u ^{4}} d u

Take the constant out:

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

= 3 (- \frac{1}{6} \cdot \int \frac{1}{u ^{4}} d u)

Apply the Power Rule

: - \frac{1}{3 u ^{3}}

= 3 (- \frac{1}{6} (- \frac{1}{3 u ^{3}}))

Substitute back

u = - 3 x^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

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