What is the integral of (2x^2-25x-33)/((x+1)^2(x-5)) ?

The integral of (2x^2-25x-33)/((x+1)^2(x-5)) is 5ln|x+1|+2ln|x-5|-29/6 ln|x-5|-1/(x+1)-1/6 ln|x-5|+C

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integral of (2x^2-25x-33)/((x+1)^2(x-5))

\int \frac{2 x ^{2} - 25 x - 33}{( x + 1 ) ^{2} ( x - 5 )} d x

5 ln ∣ x + 1 ∣ + 2 ln ∣ x - 5 ∣ - \frac{29}{6} ln ∣ x - 5 ∣ - \frac{1}{x + 1} - \frac{1}{6} ln ∣ x - 5 ∣ + C

Solution steps

\int \frac{2 x ^{2} - 25 x - 33}{( x + 1 ) ^{2} ( x - 5 )} d x

Apply u-substitution

= \int \frac{2 u ^{2} - 29 u - 6}{u ^{2} ( u - 6 )} d u

Expand

\frac{2 u ^{2} - 29 u - 6}{u ^{2} ( u - 6 )} : \frac{2 u ^{2}}{u ^{2} ( u - 6 )} - \frac{29 u}{u ^{2} ( u - 6 )} - \frac{6}{u ^{2} ( u - 6 )}

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{2 u ^{2}}{u ^{2} ( u - 6 )} d u - \int \frac{29 u}{u ^{2} ( u - 6 )} d u - \int \frac{6}{u ^{2} ( u - 6 )} d u

\int \frac{2 u ^{2}}{u ^{2} ( u - 6 )} d u = 2 ln ∣ u - 6 ∣

\int \frac{29 u}{u ^{2} ( u - 6 )} d u = 29 (- \frac{1}{6} ln ∣ u ∣ + \frac{1}{6} ln ∣ u - 6 ∣)

\int \frac{6}{u ^{2} ( u - 6 )} d u = - \frac{1}{6} ln ∣ u ∣ + \frac{1}{u} + \frac{1}{6} ln ∣ u - 6 ∣

= 2 ln ∣ u - 6 ∣ - 29 (- \frac{1}{6} ln ∣ u ∣ + \frac{1}{6} ln ∣ u - 6 ∣) - (- \frac{1}{6} ln ∣ u ∣ + \frac{1}{u} + \frac{1}{6} ln ∣ u - 6 ∣)

Substitute back

u = x + 1

= 2 ln ∣ x + 1 - 6 ∣ - 29 (- \frac{1}{6} ln ∣ x + 1 ∣ + \frac{1}{6} ln ∣ x + 1 - 6 ∣) - (- \frac{1}{6} ln ∣ x + 1 ∣ + \frac{1}{x + 1} + \frac{1}{6} ln ∣ x + 1 - 6 ∣)

Simplify

2 ln ∣ x + 1 - 6 ∣ - 29 (- \frac{1}{6} ln ∣ x + 1 ∣ + \frac{1}{6} ln ∣ x + 1 - 6 ∣) - (- \frac{1}{6} ln ∣ x + 1 ∣ + \frac{1}{x + 1} + \frac{1}{6} ln ∣ x + 1 - 6 ∣) : 5 ln ∣ x + 1 ∣ + 2 ln ∣ x - 5 ∣ - \frac{29}{6} ln ∣ x - 5 ∣ - \frac{1}{x + 1} - \frac{1}{6} ln ∣ x - 5 ∣

= 5 ln ∣ x + 1 ∣ + 2 ln ∣ x - 5 ∣ - \frac{29}{6} ln ∣ x - 5 ∣ - \frac{1}{x + 1} - \frac{1}{6} ln ∣ x - 5 ∣

Add a constant to the solution

= 5 ln ∣ x + 1 ∣ + 2 ln ∣ x - 5 ∣ - \frac{29}{6} ln ∣ x - 5 ∣ - \frac{1}{x + 1} - \frac{1}{6} ln ∣ x - 5 ∣ + C

What is the integral of (2x^2-25x-33)/((x+1)^2(x-5)) ?
The integral of (2x^2-25x-33)/((x+1)^2(x-5)) is 5ln|x+1|+2ln|x-5|-29/6 ln|x-5|-1/(x+1)-1/6 ln|x-5|+C