What is the integral of 1/(x^2(2x+1)^3) ?

The integral of 1/(x^2(2x+1)^3) is -6ln|x|-1/x+6ln|2x+1|-4/(2x+1)-1/((2x+1)^2)+C

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integral of 1/(x^2(2x+1)^3)

integral

\frac{1}{x ^{2} ( 2 x + 1 ) ^{3}}

- 6 ln ∣ x ∣ - \frac{1}{x} + 6 ln ∣ 2 x + 1 ∣ - \frac{4}{2 x + 1} - \frac{1}{( 2 x + 1 ) ^{2}} + C

Solution steps

\int \frac{1}{x ^{2} ( 2 x + 1 ) ^{3}} d x

Take the partial fraction of

\frac{1}{x ^{2} ( 2 x + 1 ) ^{3}} : - \frac{6}{x} + \frac{1}{x ^{2}} + \frac{12}{2 x + 1} + \frac{8}{( 2 x + 1 ) ^{2}} + \frac{4}{( 2 x + 1 ) ^{3}}

= \int - \frac{6}{x} + \frac{1}{x ^{2}} + \frac{12}{2 x + 1} + \frac{8}{( 2 x + 1 ) ^{2}} + \frac{4}{( 2 x + 1 ) ^{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{6}{x} d x + \int \frac{1}{x ^{2}} d x + \int \frac{12}{2 x + 1} d x + \int \frac{8}{( 2 x + 1 ) ^{2}} d x + \int \frac{4}{( 2 x + 1 ) ^{3}} d x

\int \frac{6}{x} d x = 6 ln ∣ x ∣

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

\int \frac{12}{2 x + 1} d x = 6 ln ∣ 2 x + 1 ∣

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

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

= - 6 ln ∣ x ∣ - \frac{1}{x} + 6 ln ∣ 2 x + 1 ∣ - \frac{4}{2 x + 1} - \frac{1}{( 2 x + 1 ) ^{2}}

Add a constant to the solution

= - 6 ln ∣ x ∣ - \frac{1}{x} + 6 ln ∣ 2 x + 1 ∣ - \frac{4}{2 x + 1} - \frac{1}{( 2 x + 1 ) ^{2}} + C

\frac{5}{x ^{2}}

(2 x - 1) e^{x}

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

\frac{5 x ^{2}}{( x - 2 ) ( x ^{2} + 1 )}

\frac{( x + 1 )}{( x ^{2} + x - 1 )}

What is the integral of 1/(x^2(2x+1)^3) ?
The integral of 1/(x^2(2x+1)^3) is -6ln|x|-1/x+6ln|2x+1|-4/(2x+1)-1/((2x+1)^2)+C