What is the integral of (2x^2+6x-16)/(2x+4) ?

The integral of (2x^2+6x-16)/(2x+4) is 1/2 (x+2)^2-x-2-10ln|x+2|+C

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integral of (2x^2+6x-16)/(2x+4)

\int \frac{2 x ^{2} + 6 x - 16}{2 x + 4} d x

\frac{1}{2} (x + 2)^{2} - x - 2 - 10 ln ∣ x + 2 ∣ + C

Solution steps

\int \frac{2 x ^{2} + 6 x - 16}{2 x + 4} d x

Simplify

\frac{2 x ^{2} + 6 x - 16}{2 x + 4} : \frac{x ^{2} + 3 x - 8}{x + 2}

= \int \frac{x ^{2} + 3 x - 8}{x + 2} d x

Apply u-substitution

: \int \frac{u ^{2} - u - 10}{u} d u

= \int \frac{u ^{2} - u - 10}{u} d u

Expand

\frac{u ^{2} - u - 10}{u} : u - 1 - \frac{10}{u}

= \int u - 1 - \frac{10}{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 1 d u - \int \frac{10}{u} d u

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

\int 1 d u = u

\int \frac{10}{u} d u = 10 ln ∣ u ∣

= \frac{u ^{2}}{2} - u - 10 ln ∣ u ∣

Substitute back

u = x + 2

= \frac{( x + 2 ) ^{2}}{2} - (x + 2) - 10 ln ∣ x + 2 ∣

Simplify

\frac{( x + 2 ) ^{2}}{2} - (x + 2) - 10 ln ∣ x + 2 ∣ : \frac{1}{2} (x + 2)^{2} - x - 2 - 10 ln ∣ x + 2 ∣

= \frac{1}{2} (x + 2)^{2} - x - 2 - 10 ln ∣ x + 2 ∣

Add a constant to the solution

= \frac{1}{2} (x + 2)^{2} - x - 2 - 10 ln ∣ x + 2 ∣ + C

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What is the integral of (2x^2+6x-16)/(2x+4) ?
The integral of (2x^2+6x-16)/(2x+4) is 1/2 (x+2)^2-x-2-10ln|x+2|+C