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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

Add a constant to the solution

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

\int (x - 2) d x

\int_{0}^{1} x cos (2 x^{2} + 7) d x

\frac{\partial}{\partial x} (x y^{2} + y^{2})

\int_{1}^{3} 2 r^{4} ln (r) d r

(- 4, - 4), (2, 5)

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