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

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

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

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

- \frac{1}{( x + 2 ) ^{2}} + \frac{x ^{2} ln ( \frac{1}{2} ∣ x ∣ ) - x ^{2} ln \frac{x}{2} + 1 + 4 x ln ( \frac{1}{2} ∣ x ∣ ) + 2 x - 4 x ln \frac{x}{2} + 1 + 4 ln ( \frac{1}{2} ∣ x ∣ ) - 4 ln \frac{x}{2} + 1 + 6}{8 ( x + 2 ) ^{2}} + C

Solution steps

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

Expand

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

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

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

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

= - \frac{1}{( x + 2 ) ^{2}} + \frac{x ^{2} ln ( \frac{1}{2} ∣ x ∣ ) - x ^{2} ln \frac{x}{2} + 1 + 4 x ln ( \frac{1}{2} ∣ x ∣ ) + 2 x - 4 x ln \frac{x}{2} + 1 + 4 ln ( \frac{1}{2} ∣ x ∣ ) - 4 ln \frac{x}{2} + 1 + 6}{8 ( x + 2 ) ^{2}}

Add a constant to the solution

= - \frac{1}{( x + 2 ) ^{2}} + \frac{x ^{2} ln ( \frac{1}{2} ∣ x ∣ ) - x ^{2} ln \frac{x}{2} + 1 + 4 x ln ( \frac{1}{2} ∣ x ∣ ) + 2 x - 4 x ln \frac{x}{2} + 1 + 4 ln ( \frac{1}{2} ∣ x ∣ ) - 4 ln \frac{x}{2} + 1 + 6}{8 ( x + 2 ) ^{2}} + C

2 x y d x + (x^{2} - y) d y = 0

\frac{\partial}{\partial x} (\frac{64}{x ^{2}})

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

\int (1 - x) e^{- x} d x

\frac{d}{d x} (x \cdot arctan (x))

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