What is the integral of (3x^2+15x)/((x-1)^2(x+2)) ?

The integral of (3x^2+15x)/((x-1)^2(x+2)) is 5ln|x-1|-6/(x-1)-2ln|x+2|+C

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integral of (3x^2+15x)/((x-1)^2(x+2))

\int \frac{3 x ^{2} + 15 x}{( x - 1 ) ^{2} ( x + 2 )} d x

5 ln ∣ x - 1 ∣ - \frac{6}{x - 1} - 2 ln ∣ x + 2 ∣ + C

Solution steps

\int \frac{3 x ^{2} + 15 x}{( x - 1 ) ^{2} ( x + 2 )} d x

Take the partial fraction of

\frac{3 x ^{2} + 15 x}{( x - 1 ) ^{2} ( x + 2 )} : \frac{5}{x - 1} + \frac{6}{( x - 1 ) ^{2}} - \frac{2}{x + 2}

= \int \frac{5}{x - 1} + \frac{6}{( x - 1 ) ^{2}} - \frac{2}{x + 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{5}{x - 1} d x + \int \frac{6}{( x - 1 ) ^{2}} d x - \int \frac{2}{x + 2} d x

\int \frac{5}{x - 1} d x = 5 ln ∣ x - 1 ∣

\int \frac{6}{( x - 1 ) ^{2}} d x = - \frac{6}{x - 1}

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

= 5 ln ∣ x - 1 ∣ - \frac{6}{x - 1} - 2 ln ∣ x + 2 ∣

Add a constant to the solution

= 5 ln ∣ x - 1 ∣ - \frac{6}{x - 1} - 2 ln ∣ x + 2 ∣ + C

n = 8 \sum \infty cos (\frac{1}{n ^{3}})

\frac{x - 2}{3 x ^{2} e ^{x}}

\frac{\partial}{\partial x} (3 x + 4 t)

f (x) = sin (3 ln (x))

y = 11 x + 11 x + 11 x

What is the integral of (3x^2+15x)/((x-1)^2(x+2)) ?
The integral of (3x^2+15x)/((x-1)^2(x+2)) is 5ln|x-1|-6/(x-1)-2ln|x+2|+C