What is the integral of (9x^2+5)/((4x-1)^7) ?

The integral of (9x^2+5)/((4x-1)^7) is -9/(256(4x-1)^4)-9/(160(4x-1)^5)-(89)/(384(4x-1)^6)+C

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integral of (9x^2+5)/((4x-1)^7)

\int \frac{9 x ^{2} + 5}{( 4 x - 1 ) ^{7}} d x

- \frac{9}{256 ( 4 x - 1 ) ^{4}} - \frac{9}{160 ( 4 x - 1 ) ^{5}} - \frac{89}{384 ( 4 x - 1 ) ^{6}} + C

Solution steps

\int \frac{9 x ^{2} + 5}{( 4 x - 1 ) ^{7}} d x

Take the partial fraction of

\frac{9 x ^{2} + 5}{( 4 x - 1 ) ^{7}} : \frac{9}{16 ( 4 x - 1 ) ^{5}} + \frac{9}{8 ( 4 x - 1 ) ^{6}} + \frac{89}{16 ( 4 x - 1 ) ^{7}}

= \int \frac{9}{16 ( 4 x - 1 ) ^{5}} + \frac{9}{8 ( 4 x - 1 ) ^{6}} + \frac{89}{16 ( 4 x - 1 ) ^{7}} 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{9}{16 ( 4 x - 1 ) ^{5}} d x + \int \frac{9}{8 ( 4 x - 1 ) ^{6}} d x + \int \frac{89}{16 ( 4 x - 1 ) ^{7}} d x

\int \frac{9}{16 ( 4 x - 1 ) ^{5}} d x = - \frac{9}{256 ( 4 x - 1 ) ^{4}}

\int \frac{9}{8 ( 4 x - 1 ) ^{6}} d x = - \frac{9}{160 ( 4 x - 1 ) ^{5}}

\int \frac{89}{16 ( 4 x - 1 ) ^{7}} d x = - \frac{89}{384 ( 4 x - 1 ) ^{6}}

= - \frac{9}{256 ( 4 x - 1 ) ^{4}} - \frac{9}{160 ( 4 x - 1 ) ^{5}} - \frac{89}{384 ( 4 x - 1 ) ^{6}}

Add a constant to the solution

= - \frac{9}{256 ( 4 x - 1 ) ^{4}} - \frac{9}{160 ( 4 x - 1 ) ^{5}} - \frac{89}{384 ( 4 x - 1 ) ^{6}} + C

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What is the integral of (9x^2+5)/((4x-1)^7) ?
The integral of (9x^2+5)/((4x-1)^7) is -9/(256(4x-1)^4)-9/(160(4x-1)^5)-(89)/(384(4x-1)^6)+C