What is the integral of ((4x^2+x))/((16x^3+6x^2)^2) ?

The integral of ((4x^2+x))/((16x^3+6x^2)^2) is 1/(27x)-1/(72x^2)-8/(27(8x+3))+C

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integral of ((4x^2+x))/((16x^3+6x^2)^2)

\int \frac{( 4 x ^{2} + x )}{( 16 x ^{3} + 6 x ^{2} ) ^{2}} d x

\frac{1}{27 x} - \frac{1}{72 x ^{2}} - \frac{8}{27 ( 8 x + 3 )} + C

Solution steps

\int \frac{4 x ^{2} + x}{( 16 x ^{3} + 6 x ^{2} ) ^{2}} d x

Take the partial fraction of

\frac{4 x ^{2} + x}{( 16 x ^{3} + 6 x ^{2} ) ^{2}} : - \frac{1}{27 x ^{2}} + \frac{1}{36 x ^{3}} + \frac{64}{27 ( 8 x + 3 ) ^{2}}

= \int - \frac{1}{27 x ^{2}} + \frac{1}{36 x ^{3}} + \frac{64}{27 ( 8 x + 3 ) ^{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}{27 x ^{2}} d x + \int \frac{1}{36 x ^{3}} d x + \int \frac{64}{27 ( 8 x + 3 ) ^{2}} d x

\int \frac{1}{27 x ^{2}} d x = - \frac{1}{27 x}

\int \frac{1}{36 x ^{3}} d x = - \frac{1}{72 x ^{2}}

\int \frac{64}{27 ( 8 x + 3 ) ^{2}} d x = - \frac{8}{27 ( 8 x + 3 )}

= - (- \frac{1}{27 x}) - \frac{1}{72 x ^{2}} - \frac{8}{27 ( 8 x + 3 )}

Simplify

= \frac{1}{27 x} - \frac{1}{72 x ^{2}} - \frac{8}{27 ( 8 x + 3 )}

Add a constant to the solution

= \frac{1}{27 x} - \frac{1}{72 x ^{2}} - \frac{8}{27 ( 8 x + 3 )} + C

What is the integral of ((4x^2+x))/((16x^3+6x^2)^2) ?
The integral of ((4x^2+x))/((16x^3+6x^2)^2) is 1/(27x)-1/(72x^2)-8/(27(8x+3))+C