What is the integral from 0 to 2 of ln(x+1) ?

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

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integral from 0 to 2 of ln(x+1)

\int_{0}^{2} ln (x + 1) d x

3 ln (3) - 2

Decimal Notation

1.29583 \dots

Solution steps

\int_{0}^{2} ln (x + 1) d x

Apply u-substitution

: \int_{1}^{3} ln (u) d u

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

Apply Integration By Parts

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

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

\int 1 d u = u

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

Compute the boundaries:

3 ln (3) - 2

= 3 ln (3) - 2

\int_{0}^{\frac{π}{2}} (y + y^{2} cos (x)) d x

\int_{- 6}^{0} ([\frac{1}{2} x] + ∣ x ∣) d x

\int_{- 1}^{3} ∣ x ∣ - 1 d x

\int_{0}^{5} \frac{1}{25 - t ^{2}} d t

\int_{1}^{3} \frac{1}{x ^{4}} d x

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