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

The integral of (x+2)/((x+1)^2(2x-1)) is -5/9 ln|x+1|+1/(3(x+1))+5/9 ln|2x-1|+C

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

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

- \frac{5}{9} ln ∣ x + 1 ∣ + \frac{1}{3 ( x + 1 )} + \frac{5}{9} ln ∣ 2 x - 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x + 2}{( x + 1 ) ^{2} ( 2 x - 1 )} : - \frac{5}{9 ( x + 1 )} - \frac{1}{3 ( x + 1 ) ^{2}} + \frac{10}{9 ( 2 x - 1 )}

= \int - \frac{5}{9 ( x + 1 )} - \frac{1}{3 ( x + 1 ) ^{2}} + \frac{10}{9 ( 2 x - 1 )} 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{5}{9 ( x + 1 )} d x - \int \frac{1}{3 ( x + 1 ) ^{2}} d x + \int \frac{10}{9 ( 2 x - 1 )} d x

\int \frac{5}{9 ( x + 1 )} d x = \frac{5}{9} ln ∣ x + 1 ∣

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

\int \frac{10}{9 ( 2 x - 1 )} d x = \frac{5}{9} ln ∣ 2 x - 1 ∣

= - \frac{5}{9} ln ∣ x + 1 ∣ - (- \frac{1}{3 ( x + 1 )}) + \frac{5}{9} ln ∣ 2 x - 1 ∣

Simplify

= - \frac{5}{9} ln ∣ x + 1 ∣ + \frac{1}{3 ( x + 1 )} + \frac{5}{9} ln ∣ 2 x - 1 ∣

Add a constant to the solution

= - \frac{5}{9} ln ∣ x + 1 ∣ + \frac{1}{3 ( x + 1 )} + \frac{5}{9} ln ∣ 2 x - 1 ∣ + C

\int \frac{sec ^{2} ( y )}{csc ( y )} d y

\frac{d}{d x} (\frac{3 x}{x ^{2} + 9})

x \to 0 lim (x^{2} - 2 x^{3})

n = 0 \sum \infty \frac{1}{n !} x^{n}

x \to 0 lim (- x^{2} + 4 x)

What is the integral of (x+2)/((x+1)^2(2x-1)) ?
The integral of (x+2)/((x+1)^2(2x-1)) is -5/9 ln|x+1|+1/(3(x+1))+5/9 ln|2x-1|+C