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

The integral of (3x^2+x)/((6x^3+3x^2)^2) is 1/(9x)-1/(18x^2)-2/(9(2x+1))+C

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

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

\frac{1}{9 x} - \frac{1}{18 x ^{2}} - \frac{2}{9 ( 2 x + 1 )} + C

Solution steps

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

Take the partial fraction of

\frac{3 x ^{2} + x}{( 6 x ^{3} + 3 x ^{2} ) ^{2}} : - \frac{1}{9 x ^{2}} + \frac{1}{9 x ^{3}} + \frac{4}{9 ( 2 x + 1 ) ^{2}}

= \int - \frac{1}{9 x ^{2}} + \frac{1}{9 x ^{3}} + \frac{4}{9 ( 2 x + 1 ) ^{2}} 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}{9 x ^{2}} d x + \int \frac{1}{9 x ^{3}} d x + \int \frac{4}{9 ( 2 x + 1 ) ^{2}} d x

\int \frac{1}{9 x ^{2}} d x = - \frac{1}{9 x}

\int \frac{1}{9 x ^{3}} d x = - \frac{1}{18 x ^{2}}

\int \frac{4}{9 ( 2 x + 1 ) ^{2}} d x = - \frac{2}{9 ( 2 x + 1 )}

= - (- \frac{1}{9 x}) - \frac{1}{18 x ^{2}} - \frac{2}{9 ( 2 x + 1 )}

Simplify

= \frac{1}{9 x} - \frac{1}{18 x ^{2}} - \frac{2}{9 ( 2 x + 1 )}

Add a constant to the solution

= \frac{1}{9 x} - \frac{1}{18 x ^{2}} - \frac{2}{9 ( 2 x + 1 )} + C

What is the integral of (3x^2+x)/((6x^3+3x^2)^2) ?
The integral of (3x^2+x)/((6x^3+3x^2)^2) is 1/(9x)-1/(18x^2)-2/(9(2x+1))+C