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

The integral from 1 to n of xln(x) is (n^2ln(n))/2+(-n^2+1)/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 n of xln(x)

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

\frac{n ^{2} ln ( n )}{2} + \frac{- n ^{2} + 1}{4}

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{x}{2} d x = \frac{x ^{2}}{4}

= [\frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4}]_{1}^{n}

Compute the boundaries:

\frac{1}{2} n^{2} ln (n) - \frac{n ^{2}}{4} + \frac{1}{4}

= \frac{1}{2} n^{2} ln (n) - \frac{n ^{2}}{4} + \frac{1}{4}

Simplify

= \frac{n ^{2} ln ( n )}{2} + \frac{- n ^{2} + 1}{4}

\frac{\partial}{\partial x} ((x^{2} + y) e^{\frac{y}{2}})

\int (4 - x^{2})^{\frac{3}{2}} d x

(\frac{x + 1}{x - 1})^{^{'}}

\frac{d y}{d t} = 5 y

\int x + x d x

What is the integral from 1 to n of xln(x) ?
The integral from 1 to n of xln(x) is (n^2ln(n))/2+(-n^2+1)/4