What is the integral of ((x^2-16))/((x-4)) ?

The integral of ((x^2-16))/((x-4)) is 1/2 (x-4)^2+8x-32+C

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integral of ((x^2-16))/((x-4))

\int \frac{( x ^{2} - 16 )}{( x - 4 )} d x

\frac{1}{2} (x - 4)^{2} + 8 x - 32 + C

Solution steps

\int \frac{x ^{2} - 16}{x - 4} d x

Expand

\frac{x ^{2} - 16}{x - 4} : \frac{x ^{2}}{x - 4} - \frac{16}{x - 4}

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{x ^{2}}{x - 4} d x - \int \frac{16}{x - 4} d x

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

\int \frac{16}{x - 4} d x = 16 ln ∣ x - 4 ∣

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

Simplify

\frac{( x - 4 ) ^{2}}{2} + 8 (x - 4) + 16 ln ∣ x - 4 ∣ - 16 ln ∣ x - 4 ∣ : \frac{1}{2} (x - 4)^{2} + 8 x - 32

= \frac{1}{2} (x - 4)^{2} + 8 x - 32

Add a constant to the solution

= \frac{1}{2} (x - 4)^{2} + 8 x - 32 + C

\int_{2}^{4} \frac{1}{x} d x

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

\int e^{- p x} d x

\frac{d}{d x} (x^{\frac{2}{3}} (x + 4)^{\frac{1}{2}})

d er i v a t i v e f (x) = 8 x - x^{2}

What is the integral of ((x^2-16))/((x-4)) ?
The integral of ((x^2-16))/((x-4)) is 1/2 (x-4)^2+8x-32+C