What is the integral of (6x)/((2-3x^2)^3) ?

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

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integral of (6x)/((2-3x^2)^3)

\int \frac{6 x}{( 2 - 3 x ^{2} ) ^{3}} d x

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

Solution steps

\int \frac{6 x}{( 2 - 3 x ^{2} ) ^{3}} d x

Take the constant out:

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

= 6 \cdot \int \frac{x}{( 2 - 3 x ^{2} ) ^{3}} d x

Apply u-substitution

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

Take the constant out:

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

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

Apply the Power Rule

= 6 (- \frac{1}{6} (- \frac{1}{2 u ^{2}}))

Substitute back

u = 2 - 3 x^{2}

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

Simplify

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

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

Add a constant to the solution

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

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