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

The integral of 4/(x(x-3)(x+1)^2) is -4/3 ln|x|+1/12 ln|x-3|+5/4 ln|x+1|-1/(x+1)+C

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

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

- \frac{4}{3} ln ∣ x ∣ + \frac{1}{12} ln ∣ x - 3 ∣ + \frac{5}{4} ln ∣ x + 1 ∣ - \frac{1}{x + 1} + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Take the partial fraction of

\frac{1}{x ( x - 3 ) ( x + 1 ) ^{2}} : - \frac{1}{3 x} + \frac{1}{48 ( x - 3 )} + \frac{5}{16 ( x + 1 )} + \frac{1}{4 ( x + 1 ) ^{2}}

= 4 \cdot \int - \frac{1}{3 x} + \frac{1}{48 ( x - 3 )} + \frac{5}{16 ( x + 1 )} + \frac{1}{4 ( x + 1 ) ^{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

= 4 (- \int \frac{1}{3 x} d x + \int \frac{1}{48 ( x - 3 )} d x + \int \frac{5}{16 ( x + 1 )} d x + \int \frac{1}{4 ( x + 1 ) ^{2}} d x)

\int \frac{1}{3 x} d x = \frac{1}{3} ln ∣ x ∣

\int \frac{1}{48 ( x - 3 )} d x = \frac{1}{48} ln ∣ x - 3 ∣

\int \frac{5}{16 ( x + 1 )} d x = \frac{5}{16} ln ∣ x + 1 ∣

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

= 4 (- \frac{1}{3} ln ∣ x ∣ + \frac{1}{48} ln ∣ x - 3 ∣ + \frac{5}{16} ln ∣ x + 1 ∣ - \frac{1}{4 ( x + 1 )})

Simplify

4 (- \frac{1}{3} ln ∣ x ∣ + \frac{1}{48} ln ∣ x - 3 ∣ + \frac{5}{16} ln ∣ x + 1 ∣ - \frac{1}{4 ( x + 1 )}) : - \frac{4}{3} ln ∣ x ∣ + \frac{1}{12} ln ∣ x - 3 ∣ + \frac{5}{4} ln ∣ x + 1 ∣ - \frac{1}{x + 1}

= - \frac{4}{3} ln ∣ x ∣ + \frac{1}{12} ln ∣ x - 3 ∣ + \frac{5}{4} ln ∣ x + 1 ∣ - \frac{1}{x + 1}

Add a constant to the solution

= - \frac{4}{3} ln ∣ x ∣ + \frac{1}{12} ln ∣ x - 3 ∣ + \frac{5}{4} ln ∣ x + 1 ∣ - \frac{1}{x + 1} + C

What is the integral of 4/(x(x-3)(x+1)^2) ?
The integral of 4/(x(x-3)(x+1)^2) is -4/3 ln|x|+1/12 ln|x-3|+5/4 ln|x+1|-1/(x+1)+C