What is the integral from 1 to 2 of xln(x-1) ?

The integral from 1 to 2 of xln(x-1) is -5/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 2 of xln(x-1)

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

- \frac{5}{4}

Decimal Notation

- 1.25

Solution steps

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

Apply u-substitution

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

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

Apply Integration By Parts

: [ln (u) (\frac{u ^{2}}{2} + u) - \int \frac{1}{2} (u + 2) d u]_{0}^{1}

= [ln (u) (\frac{u ^{2}}{2} + u) - \int \frac{1}{2} (u + 2) d u]_{0}^{1}

\int \frac{1}{2} (u + 2) d u = \frac{u ^{2}}{4} + u

= [ln (u) (\frac{u ^{2}}{2} + u) - (\frac{u ^{2}}{4} + u)]_{0}^{1}

Simplify

[ln (u) (\frac{u ^{2}}{2} + u) - (\frac{u ^{2}}{4} + u)]_{0}^{1} : [\frac{1}{2} u^{2} ln (u) + u ln (u) - \frac{u ^{2}}{4} - u]_{0}^{1}

= [\frac{1}{2} u^{2} ln (u) + u ln (u) - \frac{u ^{2}}{4} - u]_{0}^{1}

Compute the boundaries:

- \frac{5}{4}

= - \frac{5}{4}

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

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

\int_{0}^{3} x^{2} - 2 x + 3 d x

\int_{0}^{1} t 1 + 36 t^{2} d t

\int_{0}^{1} e^{4 x} - e^{2 x} d x

What is the integral from 1 to 2 of xln(x-1) ?
The integral from 1 to 2 of xln(x-1) is -5/4