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

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

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

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

ln ∣ x - 4 ∣ - \frac{4}{x - 4} + C

Solution steps

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

Apply u-substitution

: \int \frac{u + 4}{u ^{2}} d u

= \int \frac{u + 4}{u ^{2}} d u

Expand

\frac{u + 4}{u ^{2}} : \frac{1}{u} + \frac{4}{u ^{2}}

= \int \frac{1}{u} + \frac{4}{u ^{2}} d u

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}{u} d u + \int \frac{4}{u ^{2}} d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{4}{u ^{2}} d u = - \frac{4}{u}

= ln ∣ u ∣ - \frac{4}{u}

Substitute back

u = x - 4

= ln ∣ x - 4 ∣ - \frac{4}{x - 4}

Add a constant to the solution

= ln ∣ x - 4 ∣ - \frac{4}{x - 4} + C

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