What is the integral of x/((x-6)(x+2)^2) ?

The integral of x/((x-6)(x+2)^2) is 3/32 ln|x-6|-3/32 ln|x+2|-1/(4(x+2))+C

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integral of x/((x-6)(x+2)^2)

\int \frac{x}{( x - 6 ) ( x + 2 ) ^{2}} d x

\frac{3}{32} ln ∣ x - 6 ∣ - \frac{3}{32} ln ∣ x + 2 ∣ - \frac{1}{4 ( x + 2 )} + C

Solution steps

\int \frac{x}{( x - 6 ) ( x + 2 ) ^{2}} d x

Take the partial fraction of

\frac{x}{( x - 6 ) ( x + 2 ) ^{2}} : \frac{3}{32 ( x - 6 )} - \frac{3}{32 ( x + 2 )} + \frac{1}{4 ( x + 2 ) ^{2}}

= \int \frac{3}{32 ( x - 6 )} - \frac{3}{32 ( x + 2 )} + \frac{1}{4 ( x + 2 ) ^{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}{32 ( x - 6 )} d x - \int \frac{3}{32 ( x + 2 )} d x + \int \frac{1}{4 ( x + 2 ) ^{2}} d x

\int \frac{3}{32 ( x - 6 )} d x = \frac{3}{32} ln ∣ x - 6 ∣

\int \frac{3}{32 ( x + 2 )} d x = \frac{3}{32} ln ∣ x + 2 ∣

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

= \frac{3}{32} ln ∣ x - 6 ∣ - \frac{3}{32} ln ∣ x + 2 ∣ - \frac{1}{4 ( x + 2 )}

Add a constant to the solution

= \frac{3}{32} ln ∣ x - 6 ∣ - \frac{3}{32} ln ∣ x + 2 ∣ - \frac{1}{4 ( x + 2 )} + C

f (x) = 2 - x^{2}, g (x) = x

\int x + 1 d x

\frac{d}{d y} (\frac{y}{( x ^{2} + y ^{2} + z ^{2} ) ^{\frac{3}{2}}})

\int \frac{8 x ^{2} + 4 x + 9}{x ^{3} - 1} d x

y = x^{2}, (0, - 9)

What is the integral of x/((x-6)(x+2)^2) ?
The integral of x/((x-6)(x+2)^2) is 3/32 ln|x-6|-3/32 ln|x+2|-1/(4(x+2))+C