What is the integral of (8x+1)/((4x^2+x)^6) ?

The integral of (8x+1)/((4x^2+x)^6) is

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integral of (8x+1)/((4x^2+x)^6)

\int \frac{8 x + 1}{( 4 x ^{2} + x ) ^{6}} d x

- \frac{3584}{x} + \frac{448}{x ^{2}} - \frac{48}{x ^{3}} + \frac{4}{x ^{4}} - \frac{1}{5 x ^{5}} + \frac{14336}{4 x + 1} + \frac{7168}{( 4 x + 1 ) ^{2}} + \frac{3072}{( 4 x + 1 ) ^{3}} + \frac{1024}{( 4 x + 1 ) ^{4}} + \frac{1024}{5 ( 4 x + 1 ) ^{5}} + C

Solution steps

\int \frac{8 x + 1}{( 4 x ^{2} + x ) ^{6}} d x

Take the partial fraction of

\frac{8 x + 1}{( 4 x ^{2} + x ) ^{6}} : \frac{3584}{x ^{2}} - \frac{896}{x ^{3}} + \frac{144}{x ^{4}} - \frac{16}{x ^{5}} + \frac{1}{x ^{6}} - \frac{57344}{( 4 x + 1 ) ^{2}} - \frac{57344}{( 4 x + 1 ) ^{3}} - \frac{36864}{( 4 x + 1 ) ^{4}} - \frac{16384}{( 4 x + 1 ) ^{5}} - \frac{4096}{( 4 x + 1 ) ^{6}}

= \int \frac{3584}{x ^{2}} - \frac{896}{x ^{3}} + \frac{144}{x ^{4}} - \frac{16}{x ^{5}} + \frac{1}{x ^{6}} - \frac{57344}{( 4 x + 1 ) ^{2}} - \frac{57344}{( 4 x + 1 ) ^{3}} - \frac{36864}{( 4 x + 1 ) ^{4}} - \frac{16384}{( 4 x + 1 ) ^{5}} - \frac{4096}{( 4 x + 1 ) ^{6}} 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{3584}{x ^{2}} d x - \int \frac{896}{x ^{3}} d x + \int \frac{144}{x ^{4}} d x - \int \frac{16}{x ^{5}} d x + \int \frac{1}{x ^{6}} d x - \int \frac{57344}{( 4 x + 1 ) ^{2}} d x - \int \frac{57344}{( 4 x + 1 ) ^{3}} d x - \int \frac{36864}{( 4 x + 1 ) ^{4}} d x - \int \frac{16384}{( 4 x + 1 ) ^{5}} d x - \int \frac{4096}{( 4 x + 1 ) ^{6}} d x

\int \frac{3584}{x ^{2}} d x = - \frac{3584}{x}

\int \frac{896}{x ^{3}} d x = - \frac{448}{x ^{2}}

\int \frac{144}{x ^{4}} d x = - \frac{48}{x ^{3}}

\int \frac{16}{x ^{5}} d x = - \frac{4}{x ^{4}}

\int \frac{1}{x ^{6}} d x = - \frac{1}{5 x ^{5}}

\int \frac{57344}{( 4 x + 1 ) ^{2}} d x = - \frac{14336}{4 x + 1}

\int \frac{57344}{( 4 x + 1 ) ^{3}} d x = - \frac{7168}{( 4 x + 1 ) ^{2}}

\int \frac{36864}{( 4 x + 1 ) ^{4}} d x = - \frac{3072}{( 4 x + 1 ) ^{3}}

\int \frac{16384}{( 4 x + 1 ) ^{5}} d x = - \frac{1024}{( 4 x + 1 ) ^{4}}

\int \frac{4096}{( 4 x + 1 ) ^{6}} d x = - \frac{1024}{5 ( 4 x + 1 ) ^{5}}

= - \frac{3584}{x} - (- \frac{448}{x ^{2}}) - \frac{48}{x ^{3}} - (- \frac{4}{x ^{4}}) - \frac{1}{5 x ^{5}} - (- \frac{14336}{4 x + 1}) - (- \frac{7168}{( 4 x + 1 ) ^{2}}) - (- \frac{3072}{( 4 x + 1 ) ^{3}}) - (- \frac{1024}{( 4 x + 1 ) ^{4}}) - (- \frac{1024}{5 ( 4 x + 1 ) ^{5}})

Simplify

= - \frac{3584}{x} + \frac{448}{x ^{2}} - \frac{48}{x ^{3}} + \frac{4}{x ^{4}} - \frac{1}{5 x ^{5}} + \frac{14336}{4 x + 1} + \frac{7168}{( 4 x + 1 ) ^{2}} + \frac{3072}{( 4 x + 1 ) ^{3}} + \frac{1024}{( 4 x + 1 ) ^{4}} + \frac{1024}{5 ( 4 x + 1 ) ^{5}}

Add a constant to the solution

= - \frac{3584}{x} + \frac{448}{x ^{2}} - \frac{48}{x ^{3}} + \frac{4}{x ^{4}} - \frac{1}{5 x ^{5}} + \frac{14336}{4 x + 1} + \frac{7168}{( 4 x + 1 ) ^{2}} + \frac{3072}{( 4 x + 1 ) ^{3}} + \frac{1024}{( 4 x + 1 ) ^{4}} + \frac{1024}{5 ( 4 x + 1 ) ^{5}} + C

What is the integral of (8x+1)/((4x^2+x)^6) ?
The integral of (8x+1)/((4x^2+x)^6) is