What is the integral of (8x^3+7)/((x+1)(2x+1)^3) ?

The integral of (8x^3+7)/((x+1)(2x+1)^3) is ln|x+1|+3/(2x+1)-3/((2x+1)^2)+C

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integral of (8x^3+7)/((x+1)(2x+1)^3)

\int \frac{8 x ^{3} + 7}{( x + 1 ) ( 2 x + 1 ) ^{3}} d x

ln ∣ x + 1 ∣ + \frac{3}{2 x + 1} - \frac{3}{( 2 x + 1 ) ^{2}} + C

Solution steps

\int \frac{8 x ^{3} + 7}{( x + 1 ) ( 2 x + 1 ) ^{3}} d x

Take the partial fraction of

\frac{8 x ^{3} + 7}{( x + 1 ) ( 2 x + 1 ) ^{3}} : \frac{1}{x + 1} - \frac{6}{( 2 x + 1 ) ^{2}} + \frac{12}{( 2 x + 1 ) ^{3}}

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

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

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

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

= ln ∣ x + 1 ∣ - (- \frac{3}{2 x + 1}) - \frac{3}{( 2 x + 1 ) ^{2}}

Simplify

= ln ∣ x + 1 ∣ + \frac{3}{2 x + 1} - \frac{3}{( 2 x + 1 ) ^{2}}

Add a constant to the solution

= ln ∣ x + 1 ∣ + \frac{3}{2 x + 1} - \frac{3}{( 2 x + 1 ) ^{2}} + C

y = \frac{x ^{2}}{2} - 4 x + 6, y = \frac{x}{2} - 1

x^{\frac{1}{2}} + y^{\frac{1}{2}} = 9

\frac{d}{d x} (\frac{x + 9}{x - 9})

x \to 0 lim (\frac{( 1 + x ) ^{7} - 1}{x})

\frac{\partial}{\partial x} (ln (e^{x} + e^{y} + e^{z}))

What is the integral of (8x^3+7)/((x+1)(2x+1)^3) ?
The integral of (8x^3+7)/((x+1)(2x+1)^3) is ln|x+1|+3/(2x+1)-3/((2x+1)^2)+C