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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from t to 1 of ln(x)

\int_{t}^{1} ln (x) d x

- 1 - t ln (t) + t

Solution steps

\int_{t}^{1} ln (x) d x

Apply Integration By Parts

: [x ln (x) - \int 1 d x]_{t}^{1}

= [x ln (x) - \int 1 d x]_{t}^{1}

\int 1 d x = x

= [x ln (x) - x]_{t}^{1}

Compute the boundaries:

- 1 - t ln (t) + t

= - 1 - t ln (t) + t

\frac{1}{s ^{2} + 2}

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

\frac{1}{8 x ^{2}}

n = 1 \sum \infty 3^{n}

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

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