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

The integral of (6x+7)/((3x^2+7x)^3) is 3/(343x)-1/(98x^2)-9/(343(3x+7))-9/(98(3x+7)^2)+C

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

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

\frac{3}{343 x} - \frac{1}{98 x ^{2}} - \frac{9}{343 ( 3 x + 7 )} - \frac{9}{98 ( 3 x + 7 ) ^{2}} + C

Solution steps

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

Take the partial fraction of

\frac{6 x + 7}{( 3 x ^{2} + 7 x ) ^{3}} : - \frac{3}{343 x ^{2}} + \frac{1}{49 x ^{3}} + \frac{27}{343 ( 3 x + 7 ) ^{2}} + \frac{27}{49 ( 3 x + 7 ) ^{3}}

= \int - \frac{3}{343 x ^{2}} + \frac{1}{49 x ^{3}} + \frac{27}{343 ( 3 x + 7 ) ^{2}} + \frac{27}{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{3}{343 x ^{2}} d x + \int \frac{1}{49 x ^{3}} d x + \int \frac{27}{343 ( 3 x + 7 ) ^{2}} d x + \int \frac{27}{49 ( 3 x + 7 ) ^{3}} d x

\int \frac{3}{343 x ^{2}} d x = - \frac{3}{343 x}

\int \frac{1}{49 x ^{3}} d x = - \frac{1}{98 x ^{2}}

\int \frac{27}{343 ( 3 x + 7 ) ^{2}} d x = - \frac{9}{343 ( 3 x + 7 )}

\int \frac{27}{49 ( 3 x + 7 ) ^{3}} d x = - \frac{9}{98 ( 3 x + 7 ) ^{2}}

= - (- \frac{3}{343 x}) - \frac{1}{98 x ^{2}} - \frac{9}{343 ( 3 x + 7 )} - \frac{9}{98 ( 3 x + 7 ) ^{2}}

Simplify

= \frac{3}{343 x} - \frac{1}{98 x ^{2}} - \frac{9}{343 ( 3 x + 7 )} - \frac{9}{98 ( 3 x + 7 ) ^{2}}

Add a constant to the solution

= \frac{3}{343 x} - \frac{1}{98 x ^{2}} - \frac{9}{343 ( 3 x + 7 )} - \frac{9}{98 ( 3 x + 7 ) ^{2}} + C

\frac{5 s}{12 s ^{2} + 5 s - 2}

\frac{\partial}{\partial z} (x sin (y) + z e^{x})

\int 2 e^{- 0.1 x} d x

\frac{1}{( s ^{2} + 2 s + 2 ) ( s + 1 )}

\int 5 sin (6 π t) d t

What is the integral of (6x+7)/((3x^2+7x)^3) ?
The integral of (6x+7)/((3x^2+7x)^3) is 3/(343x)-1/(98x^2)-9/(343(3x+7))-9/(98(3x+7)^2)+C