What is the integral of (x^2+7x-6)/((x+1)(x^2-4x+4)) ?

The integral of (x^2+7x-6)/((x+1)(x^2-4x+4)) is -4/3 ln|x+1|+7/3 ln|x-2|-4/(x-2)+C

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integral of (x^2+7x-6)/((x+1)(x^2-4x+4))

\int \frac{x ^{2} + 7 x - 6}{( x + 1 ) ( x ^{2} - 4 x + 4 )} d x

- \frac{4}{3} ln ∣ x + 1 ∣ + \frac{7}{3} ln ∣ x - 2 ∣ - \frac{4}{x - 2} + C

Solution steps

\int \frac{x ^{2} + 7 x - 6}{( x + 1 ) ( x ^{2} - 4 x + 4 )} d x

Take the partial fraction of

\frac{x ^{2} + 7 x - 6}{( x + 1 ) ( x ^{2} - 4 x + 4 )} : - \frac{4}{3 ( x + 1 )} + \frac{7}{3 ( x - 2 )} + \frac{4}{( x - 2 ) ^{2}}

= \int - \frac{4}{3 ( x + 1 )} + \frac{7}{3 ( x - 2 )} + \frac{4}{( 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{4}{3 ( x + 1 )} d x + \int \frac{7}{3 ( x - 2 )} d x + \int \frac{4}{( x - 2 ) ^{2}} d x

\int \frac{4}{3 ( x + 1 )} d x = \frac{4}{3} ln ∣ x + 1 ∣

\int \frac{7}{3 ( x - 2 )} d x = \frac{7}{3} ln ∣ x - 2 ∣

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

= - \frac{4}{3} ln ∣ x + 1 ∣ + \frac{7}{3} ln ∣ x - 2 ∣ - \frac{4}{x - 2}

Add a constant to the solution

= - \frac{4}{3} ln ∣ x + 1 ∣ + \frac{7}{3} ln ∣ x - 2 ∣ - \frac{4}{x - 2} + C

What is the integral of (x^2+7x-6)/((x+1)(x^2-4x+4)) ?
The integral of (x^2+7x-6)/((x+1)(x^2-4x+4)) is -4/3 ln|x+1|+7/3 ln|x-2|-4/(x-2)+C