What is the integral of (4x^2)/((x-8)^3) ?

The integral of (4x^2)/((x-8)^3) is 4(ln|x-8|-(16)/(x-8)-(32)/((x-8)^2))+C

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integral of (4x^2)/((x-8)^3)

\int \frac{4 x ^{2}}{( x - 8 ) ^{3}} d x

4 (ln ∣ x - 8 ∣ - \frac{16}{x - 8} - \frac{32}{( x - 8 ) ^{2}}) + C

Solution steps

\int \frac{4 x ^{2}}{( x - 8 ) ^{3}} d x

Take the constant out:

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

= 4 \cdot \int \frac{x ^{2}}{( x - 8 ) ^{3}} d x

Apply u-substitution

: \int \frac{( u + 8 ) ^{2}}{u ^{3}} d u

= 4 \cdot \int \frac{( u + 8 ) ^{2}}{u ^{3}} d u

Expand

\frac{( u + 8 ) ^{2}}{u ^{3}} : \frac{1}{u} + \frac{16}{u ^{2}} + \frac{64}{u ^{3}}

= 4 \cdot \int \frac{1}{u} + \frac{16}{u ^{2}} + \frac{64}{u ^{3}} 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

= 4 (\int \frac{1}{u} d u + \int \frac{16}{u ^{2}} d u + \int \frac{64}{u ^{3}} d u)

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{16}{u ^{2}} d u = - \frac{16}{u}

\int \frac{64}{u ^{3}} d u = - \frac{32}{u ^{2}}

= 4 (ln ∣ u ∣ - \frac{16}{u} - \frac{32}{u ^{2}})

Substitute back

u = x - 8

= 4 (ln ∣ x - 8 ∣ - \frac{16}{x - 8} - \frac{32}{( x - 8 ) ^{2}})

Add a constant to the solution

= 4 (ln ∣ x - 8 ∣ - \frac{16}{x - 8} - \frac{32}{( x - 8 ) ^{2}}) + C

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What is the integral of (4x^2)/((x-8)^3) ?
The integral of (4x^2)/((x-8)^3) is 4(ln|x-8|-(16)/(x-8)-(32)/((x-8)^2))+C