What is the integral of ((2x-1)/((x+1)(x^2+4x+4))) ?

The integral of ((2x-1)/((x+1)(x^2+4x+4))) is -3ln|x+1|+3ln|x+2|-5/(x+2)+C

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integral of ((2x-1)/((x+1)(x^2+4x+4)))

\int (\frac{2 x - 1}{( x + 1 ) ( x ^{2} + 4 x + 4 )}) d x

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

Solution steps

\int \frac{2 x - 1}{( x + 1 ) ( x ^{2} + 4 x + 4 )} d x

Take the partial fraction of

\frac{2 x - 1}{( x + 1 ) ( x ^{2} + 4 x + 4 )} : - \frac{3}{x + 1} + \frac{3}{x + 2} + \frac{5}{( x + 2 ) ^{2}}

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

\int \frac{3}{x + 1} d x = 3 ln ∣ x + 1 ∣

\int \frac{3}{x + 2} d x = 3 ln ∣ x + 2 ∣

\int \frac{5}{( x + 2 ) ^{2}} d x = - \frac{5}{x + 2}

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

Add a constant to the solution

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

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What is the integral of ((2x-1)/((x+1)(x^2+4x+4))) ?
The integral of ((2x-1)/((x+1)(x^2+4x+4))) is -3ln|x+1|+3ln|x+2|-5/(x+2)+C