What is the integral of (6x+7)/((9x^2+21x)^3) ?

The integral of (6x+7)/((9x^2+21x)^3) is 1/(3087x)-1/(2646x^2)-1/(1029(3x+7))-1/(294(3x+7)^2)+C

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integral of (6x+7)/((9x^2+21x)^3)

\int \frac{6 x + 7}{( 9 x ^{2} + 21 x ) ^{3}} d x

\frac{1}{3087 x} - \frac{1}{2646 x ^{2}} - \frac{1}{1029 ( 3 x + 7 )} - \frac{1}{294 ( 3 x + 7 ) ^{2}} + C

Solution steps

\int \frac{6 x + 7}{( 9 x ^{2} + 21 x ) ^{3}} d x

Take the partial fraction of

\frac{6 x + 7}{( 9 x ^{2} + 21 x ) ^{3}} : - \frac{1}{3087 x ^{2}} + \frac{1}{1323 x ^{3}} + \frac{1}{343 ( 3 x + 7 ) ^{2}} + \frac{1}{49 ( 3 x + 7 ) ^{3}}

= \int - \frac{1}{3087 x ^{2}} + \frac{1}{1323 x ^{3}} + \frac{1}{343 ( 3 x + 7 ) ^{2}} + \frac{1}{49 ( 3 x + 7 ) ^{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}{3087 x ^{2}} d x + \int \frac{1}{1323 x ^{3}} d x + \int \frac{1}{343 ( 3 x + 7 ) ^{2}} d x + \int \frac{1}{49 ( 3 x + 7 ) ^{3}} d x

\int \frac{1}{3087 x ^{2}} d x = - \frac{1}{3087 x}

\int \frac{1}{1323 x ^{3}} d x = - \frac{1}{2646 x ^{2}}

\int \frac{1}{343 ( 3 x + 7 ) ^{2}} d x = - \frac{1}{1029 ( 3 x + 7 )}

\int \frac{1}{49 ( 3 x + 7 ) ^{3}} d x = - \frac{1}{294 ( 3 x + 7 ) ^{2}}

= - (- \frac{1}{3087 x}) - \frac{1}{2646 x ^{2}} - \frac{1}{1029 ( 3 x + 7 )} - \frac{1}{294 ( 3 x + 7 ) ^{2}}

Simplify

= \frac{1}{3087 x} - \frac{1}{2646 x ^{2}} - \frac{1}{1029 ( 3 x + 7 )} - \frac{1}{294 ( 3 x + 7 ) ^{2}}

Add a constant to the solution

= \frac{1}{3087 x} - \frac{1}{2646 x ^{2}} - \frac{1}{1029 ( 3 x + 7 )} - \frac{1}{294 ( 3 x + 7 ) ^{2}} + C

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What is the integral of (6x+7)/((9x^2+21x)^3) ?
The integral of (6x+7)/((9x^2+21x)^3) is 1/(3087x)-1/(2646x^2)-1/(1029(3x+7))-1/(294(3x+7)^2)+C