What is the integral of (4x^2-19x-27)/((x+1)^2(x-3)) ?

The integral of (4x^2-19x-27)/((x+1)^2(x-3)) is 7ln|x+1|-3ln|x-3|-1/(x+1)+C

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integral of (4x^2-19x-27)/((x+1)^2(x-3))

\int \frac{4 x ^{2} - 19 x - 27}{( x + 1 ) ^{2} ( x - 3 )} d x

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

Solution steps

\int \frac{4 x ^{2} - 19 x - 27}{( x + 1 ) ^{2} ( x - 3 )} d x

Expand

\frac{4 x ^{2} - 19 x - 27}{( x + 1 ) ^{2} ( x - 3 )} : \frac{4 x ^{2}}{( x + 1 ) ^{2} ( x - 3 )} - \frac{19 x}{( x + 1 ) ^{2} ( x - 3 )} - \frac{27}{( x + 1 ) ^{2} ( x - 3 )}

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 x ^{2}}{( x + 1 ) ^{2} ( x - 3 )} d x - \int \frac{19 x}{( x + 1 ) ^{2} ( x - 3 )} d x - \int \frac{27}{( x + 1 ) ^{2} ( x - 3 )} d x

\int \frac{4 x ^{2}}{( x + 1 ) ^{2} ( x - 3 )} d x = \frac{7}{4} ln ∣ x + 1 ∣ + \frac{1}{x + 1} + \frac{9}{4} ln ∣ x - 3 ∣

\int \frac{19 x}{( x + 1 ) ^{2} ( x - 3 )} d x = 19 (- \frac{3}{16} ln ∣ x + 1 ∣ - \frac{1}{4 ( x + 1 )} + \frac{3}{16} ln ∣ x - 3 ∣)

\int \frac{27}{( x + 1 ) ^{2} ( x - 3 )} d x = 27 (- \frac{1}{16} ln ∣ x + 1 ∣ + \frac{1}{4 ( x + 1 )} + \frac{1}{16} ln ∣ x - 3 ∣)

= \frac{7}{4} ln ∣ x + 1 ∣ + \frac{1}{x + 1} + \frac{9}{4} ln ∣ x - 3 ∣ - 19 (- \frac{3}{16} ln ∣ x + 1 ∣ - \frac{1}{4 ( x + 1 )} + \frac{3}{16} ln ∣ x - 3 ∣) - 27 (- \frac{1}{16} ln ∣ x + 1 ∣ + \frac{1}{4 ( x + 1 )} + \frac{1}{16} ln ∣ x - 3 ∣)

Simplify

\frac{7}{4} ln ∣ x + 1 ∣ + \frac{1}{x + 1} + \frac{9}{4} ln ∣ x - 3 ∣ - 19 (- \frac{3}{16} ln ∣ x + 1 ∣ - \frac{1}{4 ( x + 1 )} + \frac{3}{16} ln ∣ x - 3 ∣) - 27 (- \frac{1}{16} ln ∣ x + 1 ∣ + \frac{1}{4 ( x + 1 )} + \frac{1}{16} ln ∣ x - 3 ∣) : 7 ln ∣ x + 1 ∣ - 3 ln ∣ x - 3 ∣ - \frac{1}{x + 1}

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

Add a constant to the solution

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

lo g_{4} (x)

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What is the integral of (4x^2-19x-27)/((x+1)^2(x-3)) ?
The integral of (4x^2-19x-27)/((x+1)^2(x-3)) is 7ln|x+1|-3ln|x-3|-1/(x+1)+C