What is the integral of (x^2)/(2x+12) ?

The integral of (x^2)/(2x+12) is 1/4 (72ln(2|x+6|)+(x+6)(x-18))+C

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integral of (x^2)/(2x+12)

\int \frac{x ^{2}}{2 x + 12} d x

\frac{1}{4} (72 ln (2 ∣ x + 6 ∣) + (x + 6) (x - 18)) + C

Solution steps

\int \frac{x ^{2}}{2 x + 12} d x

Apply u-substitution

= \int \frac{( u - 12 ) ^{2}}{8 u} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{( u - 12 ) ^{2}}{u} d u

Expand

\frac{( u - 12 ) ^{2}}{u} : u - 24 + \frac{144}{u}

= \frac{1}{8} \cdot \int u - 24 + \frac{144}{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

= \frac{1}{8} (\int u d u - \int 24 d u + \int \frac{144}{u} d u)

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

\int 24 d u = 24 u

\int \frac{144}{u} d u = 144 ln ∣ u ∣

= \frac{1}{8} (\frac{u ^{2}}{2} - 24 u + 144 ln ∣ u ∣)

Substitute back

u = 2 x + 12

= \frac{1}{8} (\frac{( 2 x + 12 ) ^{2}}{2} - 24 (2 x + 12) + 144 ln ∣ 2 x + 12 ∣)

Simplify

\frac{1}{8} (\frac{( 2 x + 12 ) ^{2}}{2} - 24 (2 x + 12) + 144 ln ∣ 2 x + 12 ∣) : \frac{1}{4} (72 ln (2 ∣ x + 6 ∣) + (x + 6) (x - 18))

= \frac{1}{4} (72 ln (2 ∣ x + 6 ∣) + (x + 6) (x - 18))

Add a constant to the solution

= \frac{1}{4} (72 ln (2 ∣ x + 6 ∣) + (x + 6) (x - 18)) + C

What is the integral of (x^2)/(2x+12) ?
The integral of (x^2)/(2x+12) is 1/4 (72ln(2|x+6|)+(x+6)(x-18))+C