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

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

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

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

ln ∣ x + 1 ∣ - ln ∣ 2 x + 1 ∣ - \frac{1}{2 x + 1} + C

Solution steps

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

Take the partial fraction of

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

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

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

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

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

= ln ∣ x + 1 ∣ - ln ∣ 2 x + 1 ∣ - \frac{1}{2 x + 1}

Add a constant to the solution

= ln ∣ x + 1 ∣ - ln ∣ 2 x + 1 ∣ - \frac{1}{2 x + 1} + C

x^{3} - 3 x^{2} - 9 x + 11, at x = 3

\int (cos (θ))^{2} d θ

\int (ln (t))^{2} d t

x \to 6 lim (x^{3} + 3 x)

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

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