What is the integral of 1/12 ln^2(3x+1) ?

The integral of 1/12 ln^2(3x+1) is 1/36 (ln^2(3x+1)(3x+1)-2(3xln(3x+1)+ln(3x+1)-3x-1))+C

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integral of 1/12 ln^2(3x+1)

\int \frac{1}{12} ln^{2} (3 x + 1) d x

\frac{1}{36} (ln^{2} (3 x + 1) (3 x + 1) - 2 (3 x ln (3 x + 1) + ln (3 x + 1) - 3 x - 1)) + C

Solution steps

\int \frac{1}{12} ln^{2} (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}{12} \cdot \int ln^{2} (3 x + 1) d x

Apply u-substitution

: \int ln^{2} (u) \frac{1}{3} d u

= \frac{1}{12} \cdot \int ln^{2} (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}{12} \cdot \frac{1}{3} \cdot \int ln^{2} (u) d u

Apply Integration By Parts

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

= \frac{1}{12} \cdot \frac{1}{3} (u ln^{2} (u) - \int 2 ln (u) d u)

\int 2 ln (u) d u = 2 (u ln (u) - u)

= \frac{1}{12} \cdot \frac{1}{3} (u ln^{2} (u) - 2 (u ln (u) - u))

Substitute back

u = 3 x + 1

= \frac{1}{12} \cdot \frac{1}{3} ((3 x + 1) ln^{2} (3 x + 1) - 2 ((3 x + 1) ln (3 x + 1) - (3 x + 1)))

Simplify

\frac{1}{12} \cdot \frac{1}{3} ((3 x + 1) ln^{2} (3 x + 1) - 2 ((3 x + 1) ln (3 x + 1) - (3 x + 1))) : \frac{1}{36} (ln^{2} (3 x + 1) (3 x + 1) - 2 (3 x ln (3 x + 1) + ln (3 x + 1) - 3 x - 1))

= \frac{1}{36} (ln^{2} (3 x + 1) (3 x + 1) - 2 (3 x ln (3 x + 1) + ln (3 x + 1) - 3 x - 1))

Add a constant to the solution

= \frac{1}{36} (ln^{2} (3 x + 1) (3 x + 1) - 2 (3 x ln (3 x + 1) + ln (3 x + 1) - 3 x - 1)) + C

\int \frac{1}{2 + 2 x ^{2}} d x

1 - \frac{1}{t}

\int \frac{8}{8 + e ^{x}} d x

y^{2} = 9 x, at x = 16

What is the integral of 1/12 ln^2(3x+1) ?
The integral of 1/12 ln^2(3x+1) is 1/36 (ln^2(3x+1)(3x+1)-2(3xln(3x+1)+ln(3x+1)-3x-1))+C