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

The integral of (x^2)/(x-8) is ((x-8)^2)/2+16(x-8)+64ln|x-8|+C

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

\int \frac{x ^{2}}{x - 8} d x

\frac{( x - 8 ) ^{2}}{2} + 16 (x - 8) + 64 ln ∣ x - 8 ∣ + C

Solution steps

\int \frac{x ^{2}}{x - 8} d x

Apply u-substitution

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

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

Expand

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

= \int u + 16 + \frac{64}{u} 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 u d u + \int 16 d u + \int \frac{64}{u} d u

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

\int 16 d u = 16 u

\int \frac{64}{u} d u = 64 ln ∣ u ∣

= \frac{u ^{2}}{2} + 16 u + 64 ln ∣ u ∣

Substitute back

u = x - 8

= \frac{( x - 8 ) ^{2}}{2} + 16 (x - 8) + 64 ln ∣ x - 8 ∣

Add a constant to the solution

= \frac{( x - 8 ) ^{2}}{2} + 16 (x - 8) + 64 ln ∣ x - 8 ∣ + C

\int_{- \infty}^{0} \frac{1}{x ^{2}} d x

e^{- y} (1 + \frac{d y}{d x}) = 1

\int (t - 1) e^{- s t} d t

- y = x y^{^{'}}

x y^{^{'}} + (x + 1) y = 6

What is the integral of (x^2)/(x-8) ?
The integral of (x^2)/(x-8) is ((x-8)^2)/2+16(x-8)+64ln|x-8|+C