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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

2 ln (2) - 1

Decimal

0.38629 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{2} ln (u) d u

= \int_{1}^{2} ln (u) d u

Apply Integration By Parts

: [u ln (u) - \int 1 d u]_{1}^{2}

= [u ln (u) - \int 1 d u]_{1}^{2}

\int 1 d u = u

= [u ln (u) - u]_{1}^{2}

Compute the boundaries:

2 ln (2) - 1

= 2 ln (2) - 1

\int_{2}^{3} (3 x^{2} + 2 x) d x

\int_{1}^{6} (\frac{3 x ^{2} + 2 x + 1}{x}) d x

\int_{0}^{\frac{π}{4}} 2 cos^{2} (x) d x

\int_{- 3}^{3} cos^{2} (\frac{π x}{2}) d x

\int_{x}^{y} (2 t - 3) d t

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