What is the integral of 1/((3t+6)ln(t+2)) ?

The integral of 1/((3t+6)ln(t+2)) is 1/3 ln|ln(t+2)|+C

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integral of 1/((3t+6)ln(t+2))

\int \frac{1}{( 3 t + 6 ) ln ( t + 2 )} d t

\frac{1}{3} ln ∣ ln (t + 2) ∣ + C

Solution steps

\int \frac{1}{( 3 t + 6 ) ln ( t + 2 )} d t

\frac{1}{( 3 t + 6 ) ln ( t + 2 )} = \frac{1}{3} \frac{1}{( t + 2 ) ln ( t + 2 )}

= \int \frac{1}{3} \cdot \frac{1}{( t + 2 ) ln ( t + 2 )} d t

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{( t + 2 ) ln ( t + 2 )} d t

Apply u-substitution

: \int \frac{1}{u ln ( u )} d u

= \frac{1}{3} \cdot \int \frac{1}{u ln ( u )} d u

Apply u-substitution

: \int \frac{1}{v} d v

= \frac{1}{3} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{3} ln ∣ v ∣

Substitute back

= \frac{1}{3} ln ∣ ln (t + 2) ∣

Add a constant to the solution

= \frac{1}{3} ln ∣ ln (t + 2) ∣ + C

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What is the integral of 1/((3t+6)ln(t+2)) ?
The integral of 1/((3t+6)ln(t+2)) is 1/3 ln|ln(t+2)|+C