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

The integral of 1/2 ln(3x+1) is 1/6 (3xln(3x+1)+ln(3x+1)-3x-1)+C

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integral of 1/2 ln(3x+1)

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

\frac{1}{6} (3 x ln (3 x + 1) + ln (3 x + 1) - 3 x - 1) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integration By Parts

: u ln (u) - \int 1 d u

= \frac{1}{2} \cdot \frac{1}{3} (u ln (u) - \int 1 d u)

\int 1 d u = u

= \frac{1}{2} \cdot \frac{1}{3} (u ln (u) - u)

Substitute back

u = 3 x + 1

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of 1/2 ln(3x+1) ?
The integral of 1/2 ln(3x+1) is 1/6 (3xln(3x+1)+ln(3x+1)-3x-1)+C