What is the integral of (4x^2-12x-19)/((x+1)^2(x-2)) ?

The integral of (4x^2-12x-19)/((x+1)^2(x-2)) is 7ln|x+1|-1/(x+1)-3ln|x-2|+C

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integral of (4x^2-12x-19)/((x+1)^2(x-2))

\int \frac{4 x ^{2} - 12 x - 19}{( x + 1 ) ^{2} ( x - 2 )} d x

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

Solution steps

\int \frac{4 x ^{2} - 12 x - 19}{( x + 1 ) ^{2} ( x - 2 )} d x

Take the partial fraction of

\frac{4 x ^{2} - 12 x - 19}{( x + 1 ) ^{2} ( x - 2 )} : \frac{7}{x + 1} + \frac{1}{( x + 1 ) ^{2}} - \frac{3}{x - 2}

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

\int \frac{7}{x + 1} d x = 7 ln ∣ x + 1 ∣

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

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

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

Add a constant to the solution

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

\frac{d y}{d x} (x + 3) = y

\int e^{2 y} d y

\frac{d y}{d x} = (3 x^{2} \cdot (1 + y^{2}))

y = x^{2} + 9

\frac{x - x}{x ^{\frac{1}{4}}}

What is the integral of (4x^2-12x-19)/((x+1)^2(x-2)) ?
The integral of (4x^2-12x-19)/((x+1)^2(x-2)) is 7ln|x+1|-1/(x+1)-3ln|x-2|+C