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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{16 x} d x = \frac{1}{16} ln ∣ x ∣

\int \frac{1}{16 ( x + 4 )} d x = \frac{1}{16} ln ∣ x + 4 ∣

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

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

Simplify

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

Add a constant to the solution

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

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