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

The integral of x/((x-3)(x+1)^2) is 3/16 ln|x-3|-3/16 ln|x+1|-1/(4(x+1))+C

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

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

\frac{3}{16} ln ∣ x - 3 ∣ - \frac{3}{16} ln ∣ x + 1 ∣ - \frac{1}{4 ( x + 1 )} + C

Solution steps

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

Take the partial fraction of

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

= \int \frac{3}{16 ( x - 3 )} - \frac{3}{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

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

\int \frac{3}{16 ( x - 3 )} d x = \frac{3}{16} ln ∣ x - 3 ∣

\int \frac{3}{16 ( x + 1 )} d x = \frac{3}{16} ln ∣ x + 1 ∣

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

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

Add a constant to the solution

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

What is the integral of x/((x-3)(x+1)^2) ?
The integral of x/((x-3)(x+1)^2) is 3/16 ln|x-3|-3/16 ln|x+1|-1/(4(x+1))+C