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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

- 1

Solution steps

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

Apply u-substitution

: \int_{0}^{1} ln (u) d u

= \int_{0}^{1} ln (u) d u

Apply Integration By Parts

: [u ln (u) - \int 1 d u]_{0}^{1}

= [u ln (u) - \int 1 d u]_{0}^{1}

\int 1 d u = u

= [u ln (u) - u]_{0}^{1}

Compute the boundaries:

- 1

= - 1

\int_{- 1}^{1} x^{3} (3 + x^{4})^{3} d x

\int_{- \infty}^{0} \frac{1}{x - 1} d x

\int_{0}^{π} sin^{6} (\frac{x}{2}) d x

\int_{0}^{a} sin^{2} (k x) d x

\int_{1}^{2} 2 x^{2} + x d x

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