What is the integral of (4x^2-x-1)/((x+1)^2(x-3)) ?

The integral of (4x^2-x-1)/((x+1)^2(x-3)) is 2ln|x+1|+4ln|x-3|-9/4 ln|x-3|+1/(x+1)+1/4 ln|x-3|+C

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integral of (4x^2-x-1)/((x+1)^2(x-3))

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

2 ln ∣ x + 1 ∣ + 4 ln ∣ x - 3 ∣ - \frac{9}{4} ln ∣ x - 3 ∣ + \frac{1}{x + 1} + \frac{1}{4} ln ∣ x - 3 ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{4 u ^{2} - 9 u + 4}{u ^{2} ( u - 4 )} d u

= \int \frac{4 u ^{2} - 9 u + 4}{u ^{2} ( u - 4 )} d u

Expand

\frac{4 u ^{2} - 9 u + 4}{u ^{2} ( u - 4 )} : \frac{4 u ^{2}}{u ^{2} ( u - 4 )} - \frac{9 u}{u ^{2} ( u - 4 )} + \frac{4}{u ^{2} ( u - 4 )}

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 u ^{2}}{u ^{2} ( u - 4 )} d u - \int \frac{9 u}{u ^{2} ( u - 4 )} d u + \int \frac{4}{u ^{2} ( u - 4 )} d u

\int \frac{4 u ^{2}}{u ^{2} ( u - 4 )} d u = 4 ln ∣ u - 4 ∣

\int \frac{9 u}{u ^{2} ( u - 4 )} d u = 9 (- \frac{1}{4} ln ∣ u ∣ + \frac{1}{4} ln ∣ u - 4 ∣)

\int \frac{4}{u ^{2} ( u - 4 )} d u = - \frac{1}{4} ln ∣ u ∣ + \frac{1}{u} + \frac{1}{4} ln ∣ u - 4 ∣

= 4 ln ∣ u - 4 ∣ - 9 (- \frac{1}{4} ln ∣ u ∣ + \frac{1}{4} ln ∣ u - 4 ∣) - \frac{1}{4} ln ∣ u ∣ + \frac{1}{u} + \frac{1}{4} ln ∣ u - 4 ∣

Substitute back

u = x + 1

= 4 ln ∣ x + 1 - 4 ∣ - 9 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x + 1 - 4 ∣) - \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{x + 1} + \frac{1}{4} ln ∣ x + 1 - 4 ∣

Simplify

4 ln ∣ x + 1 - 4 ∣ - 9 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x + 1 - 4 ∣) - \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{x + 1} + \frac{1}{4} ln ∣ x + 1 - 4 ∣ : 2 ln ∣ x + 1 ∣ + 4 ln ∣ x - 3 ∣ - \frac{9}{4} ln ∣ x - 3 ∣ + \frac{1}{x + 1} + \frac{1}{4} ln ∣ x - 3 ∣

= 2 ln ∣ x + 1 ∣ + 4 ln ∣ x - 3 ∣ - \frac{9}{4} ln ∣ x - 3 ∣ + \frac{1}{x + 1} + \frac{1}{4} ln ∣ x - 3 ∣

Add a constant to the solution

= 2 ln ∣ x + 1 ∣ + 4 ln ∣ x - 3 ∣ - \frac{9}{4} ln ∣ x - 3 ∣ + \frac{1}{x + 1} + \frac{1}{4} ln ∣ x - 3 ∣ + C

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What is the integral of (4x^2-x-1)/((x+1)^2(x-3)) ?
The integral of (4x^2-x-1)/((x+1)^2(x-3)) is 2ln|x+1|+4ln|x-3|-9/4 ln|x-3|+1/(x+1)+1/4 ln|x-3|+C