What is the integral of-4/(x(x-396)) ?

The integral of-4/(x(x-396)) is -4(-1/396 ln|x|+1/396 ln|x-396|)+C

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integral of-4/(x(x-396))

\int - \frac{4}{x ( x - 396 )} d x

- 4 (- \frac{1}{396} ln ∣ x ∣ + \frac{1}{396} ln ∣ x - 396 ∣) + C

Solution steps

\int - \frac{4}{x ( x - 396 )} 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 - 396 )} d x

Take the partial fraction of

\frac{1}{x ( x - 396 )} : - \frac{1}{396 x} + \frac{1}{396 ( x - 396 )}

= - 4 \cdot \int - \frac{1}{396 x} + \frac{1}{396 ( x - 396 )} 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}{396 x} d x + \int \frac{1}{396 ( x - 396 )} d x)

\int \frac{1}{396 x} d x = \frac{1}{396} ln ∣ x ∣

\int \frac{1}{396 ( x - 396 )} d x = \frac{1}{396} ln ∣ x - 396 ∣

= - 4 (- \frac{1}{396} ln ∣ x ∣ + \frac{1}{396} ln ∣ x - 396 ∣)

Add a constant to the solution

= - 4 (- \frac{1}{396} ln ∣ x ∣ + \frac{1}{396} ln ∣ x - 396 ∣) + C

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What is the integral of-4/(x(x-396)) ?
The integral of-4/(x(x-396)) is -4(-1/396 ln|x|+1/396 ln|x-396|)+C