What is the integral from 0 to 1 of ln(1+3x) ?

The integral from 0 to 1 of ln(1+3x) is 8/3 ln(2)-1

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integral from 0 to 1 of ln(1+3x)

\int_{0}^{1} ln (1 + 3 x) d x

\frac{8}{3} ln (2) - 1

Decimal

0.84839 \dots

Solution steps

\int_{0}^{1} ln (1 + 3 x) d x

Apply u-substitution

= \int_{1}^{4} 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}{3} \cdot \int_{1}^{4} ln (u) d u

Apply Integration By Parts

= \frac{1}{3} [u ln (u) - \int 1 d u]_{1}^{4}

\int 1 d u = u

= \frac{1}{3} [u ln (u) - u]_{1}^{4}

Compute the boundaries:

8 ln (2) - 3

= \frac{1}{3} (8 ln (2) - 3)

Simplify

= \frac{8}{3} ln (2) - 1

What is the integral from 0 to 1 of ln(1+3x) ?
The integral from 0 to 1 of ln(1+3x) is 8/3 ln(2)-1