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

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

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

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

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

Solution steps

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

Apply u-substitution

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

Expand

\frac{( u + 4 ) ^{2}}{u} : u + 8 + \frac{16}{u}

= \int u + 8 + \frac{16}{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 8 d u + \int \frac{16}{u} d u

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

\int 8 d u = 8 u

\int \frac{16}{u} d u = 16 ln ∣ u ∣

= \frac{u ^{2}}{2} + 8 u + 16 ln ∣ u ∣

Substitute back

u = x - 4

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

Add a constant to the solution

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

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