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

The integral of (3x^2+2)/((3x^3+6x)^3) is -1/(216x^2)+1/(216(x^2+2))+1/(108(x^2+2)^2)+C

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

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

- \frac{1}{216 x ^{2}} + \frac{1}{216 ( x ^{2} + 2 )} + \frac{1}{108 ( x ^{2} + 2 ) ^{2}} + C

Solution steps

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

Take the partial fraction of

\frac{3 x ^{2} + 2}{( 3 x ^{3} + 6 x ) ^{3}} : \frac{1}{108 x ^{3}} - \frac{x}{108 ( x ^{2} + 2 ) ^{2}} - \frac{x}{27 ( x ^{2} + 2 ) ^{3}}

= \int \frac{1}{108 x ^{3}} - \frac{x}{108 ( x ^{2} + 2 ) ^{2}} - \frac{x}{27 ( x ^{2} + 2 ) ^{3}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{108 x ^{3}} d x - \int \frac{x}{108 ( x ^{2} + 2 ) ^{2}} d x - \int \frac{x}{27 ( x ^{2} + 2 ) ^{3}} d x

\int \frac{1}{108 x ^{3}} d x = - \frac{1}{216 x ^{2}}

\int \frac{x}{108 ( x ^{2} + 2 ) ^{2}} d x = - \frac{1}{216 ( x ^{2} + 2 )}

\int \frac{x}{27 ( x ^{2} + 2 ) ^{3}} d x = - \frac{1}{108 ( x ^{2} + 2 ) ^{2}}

= - \frac{1}{216 x ^{2}} - (- \frac{1}{216 ( x ^{2} + 2 )}) - (- \frac{1}{108 ( x ^{2} + 2 ) ^{2}})

Simplify

= - \frac{1}{216 x ^{2}} + \frac{1}{216 ( x ^{2} + 2 )} + \frac{1}{108 ( x ^{2} + 2 ) ^{2}}

Add a constant to the solution

= - \frac{1}{216 x ^{2}} + \frac{1}{216 ( x ^{2} + 2 )} + \frac{1}{108 ( x ^{2} + 2 ) ^{2}} + C

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What is the integral of (3x^2+2)/((3x^3+6x)^3) ?
The integral of (3x^2+2)/((3x^3+6x)^3) is -1/(216x^2)+1/(216(x^2+2))+1/(108(x^2+2)^2)+C