What is the integral of (5x^2)/(x(x-4)^2) ?

The integral of (5x^2)/(x(x-4)^2) is 5(ln|x-4|-4/(x-4))+C

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integral of (5x^2)/(x(x-4)^2)

\int \frac{5 x ^{2}}{x ( x - 4 ) ^{2}} d x

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

Solution steps

\int \frac{5 x ^{2}}{x ( x - 4 ) ^{2}} d x

Take the constant out:

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

= 5 \cdot \int \frac{x ^{2}}{x ( x - 4 ) ^{2}} d x

Cancel the common factor:

x

= 5 \cdot \int \frac{x}{( x - 4 ) ^{2}} d x

Apply u-substitution

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

= 5 \cdot \int \frac{u + 4}{u ^{2}} d u

Expand

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

= 5 \cdot \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

= 5 (\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}

= 5 (ln ∣ u ∣ - \frac{4}{u})

Substitute back

u = x - 4

= 5 (ln ∣ x - 4 ∣ - \frac{4}{x - 4})

Add a constant to the solution

= 5 (ln ∣ x - 4 ∣ - \frac{4}{x - 4}) + C

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What is the integral of (5x^2)/(x(x-4)^2) ?
The integral of (5x^2)/(x(x-4)^2) is 5(ln|x-4|-4/(x-4))+C