What is the integral of 7/(x(x-7)^2) ?

The integral of 7/(x(x-7)^2) is 1/7 ln|x|-1/7 ln|x-7|-1/(x-7)+C

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integral of 7/(x(x-7)^2)

\int \frac{7}{x ( x - 7 ) ^{2}} d x

\frac{1}{7} ln ∣ x ∣ - \frac{1}{7} ln ∣ x - 7 ∣ - \frac{1}{x - 7} + C

Solution steps

\int \frac{7}{x ( x - 7 ) ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 7 \cdot \int \frac{1}{x ( x - 7 ) ^{2}} d x

Take the partial fraction of

\frac{1}{x ( x - 7 ) ^{2}} : \frac{1}{49 x} - \frac{1}{49 ( x - 7 )} + \frac{1}{7 ( x - 7 ) ^{2}}

= 7 \cdot \int \frac{1}{49 x} - \frac{1}{49 ( x - 7 )} + \frac{1}{7 ( x - 7 ) ^{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

= 7 (\int \frac{1}{49 x} d x - \int \frac{1}{49 ( x - 7 )} d x + \int \frac{1}{7 ( x - 7 ) ^{2}} d x)

\int \frac{1}{49 x} d x = \frac{1}{49} ln ∣ x ∣

\int \frac{1}{49 ( x - 7 )} d x = \frac{1}{49} ln ∣ x - 7 ∣

\int \frac{1}{7 ( x - 7 ) ^{2}} d x = - \frac{1}{7 ( x - 7 )}

= 7 (\frac{1}{49} ln ∣ x ∣ - \frac{1}{49} ln ∣ x - 7 ∣ - \frac{1}{7 ( x - 7 )})

Simplify

7 (\frac{1}{49} ln ∣ x ∣ - \frac{1}{49} ln ∣ x - 7 ∣ - \frac{1}{7 ( x - 7 )}) : \frac{1}{7} ln ∣ x ∣ - \frac{1}{7} ln ∣ x - 7 ∣ - \frac{1}{x - 7}

= \frac{1}{7} ln ∣ x ∣ - \frac{1}{7} ln ∣ x - 7 ∣ - \frac{1}{x - 7}

Add a constant to the solution

= \frac{1}{7} ln ∣ x ∣ - \frac{1}{7} ln ∣ x - 7 ∣ - \frac{1}{x - 7} + C

What is the integral of 7/(x(x-7)^2) ?
The integral of 7/(x(x-7)^2) is 1/7 ln|x|-1/7 ln|x-7|-1/(x-7)+C