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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

Add a constant to the solution

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

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