What is the integral of ln(6x+6x^2) ?

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

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integral of ln(6x+6x^2)

\int ln (6 x + 6 x^{2}) d x

x ln (6 x^{2} + 6 x) + \frac{1}{2} ln (6 x^{2} + 6 x) + \frac{1}{2} ln ∣ 2 x + 2 ∣ - \frac{1}{2} ln (2 ∣ x ∣) - 2 x - 1 + C

Solution steps

\int ln (6 x + 6 x^{2}) d x

Complete the square

6 x + 6 x^{2} : 6 (x + \frac{1}{2})^{2} - \frac{3}{2}

= \int ln (6 (x + \frac{1}{2})^{2} - \frac{3}{2}) d x

Apply u-substitution

: \int ln (6 u^{2} - \frac{3}{2}) d u

= \int ln (6 u^{2} - \frac{3}{2}) d u

Apply Integration By Parts

: u ln (6 u^{2} - \frac{3}{2}) - \int \frac{8 u ^{2}}{4 u ^{2} - 1} d u

= u ln (6 u^{2} - \frac{3}{2}) - \int \frac{8 u ^{2}}{4 u ^{2} - 1} d u

\int \frac{8 u ^{2}}{4 u ^{2} - 1} d u = - \frac{1}{2} ln ∣ 2 u + 1 ∣ + \frac{1}{2} ln ∣ 2 u - 1 ∣ + 2 u

= u ln (6 u^{2} - \frac{3}{2}) - (- \frac{1}{2} ln ∣ 2 u + 1 ∣ + \frac{1}{2} ln ∣ 2 u - 1 ∣ + 2 u)

Substitute back

u = x + \frac{1}{2}

= (x + \frac{1}{2}) ln (6 (x + \frac{1}{2})^{2} - \frac{3}{2}) - (- \frac{1}{2} ln 2 (x + \frac{1}{2}) + 1 + \frac{1}{2} ln 2 (x + \frac{1}{2}) - 1 + 2 (x + \frac{1}{2}))

Simplify

(x + \frac{1}{2}) ln (6 (x + \frac{1}{2})^{2} - \frac{3}{2}) - (- \frac{1}{2} ln 2 (x + \frac{1}{2}) + 1 + \frac{1}{2} ln 2 (x + \frac{1}{2}) - 1 + 2 (x + \frac{1}{2})) : x ln (6 x^{2} + 6 x) + \frac{1}{2} ln (6 x^{2} + 6 x) + \frac{1}{2} ln ∣ 2 x + 2 ∣ - \frac{1}{2} ln (2 ∣ x ∣) - 2 x - 1

= x ln (6 x^{2} + 6 x) + \frac{1}{2} ln (6 x^{2} + 6 x) + \frac{1}{2} ln ∣ 2 x + 2 ∣ - \frac{1}{2} ln (2 ∣ x ∣) - 2 x - 1

Add a constant to the solution

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

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