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

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

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integral of 4/(x(x-4)^2)

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

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

Solution steps

\int \frac{4}{x ( x - 4 ) ^{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 - 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}}

= 4 \cdot \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

= 4 (\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 )}

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

Simplify

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

= \frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x - 4 ∣ - \frac{1}{x - 4}

Add a constant to the solution

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

x \to 1 + lim (arcsin (x))

\frac{d}{d x} (ln (ln (ln (2 x))))

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

x \to \infty lim (x \cdot e^{x})

\frac{d}{d x} (e^{2 x} (3 x - 2))

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