What is the integral of (ln(x+2))/(2x+4) ?

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

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integral of (ln(x+2))/(2x+4)

\int \frac{ln ( x + 2 )}{2 x + 4} d x

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

Solution steps

\int \frac{ln ( x + 2 )}{2 x + 4} d x

Apply Integration By Parts

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

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

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

Simplify

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

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

Add a constant to the solution

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

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