What is the integral from 1 to e of sqrt(x)ln(x^2) ?

The integral from 1 to e of sqrt(x)ln(x^2) is (4e^{3/2}+8)/9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to e of sqrt(x)ln(x^2)

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

\frac{4 e ^{\frac{3}{2}} + 8}{9}

Decimal Notation

2.88075 \dots

Solution steps

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

Apply Integration By Parts

: [\frac{1}{3} (2 x^{\frac{3}{2}} ln (x^{2}) - 3 \cdot \int \frac{4 x ^{\frac{1}{2}}}{3} d x)]_{1}^{e}

= [\frac{1}{3} (2 x^{\frac{3}{2}} ln (x^{2}) - 3 \cdot \int \frac{4 x ^{\frac{1}{2}}}{3} d x)]_{1}^{e}

\int \frac{4 x ^{\frac{1}{2}}}{3} d x = \frac{8}{9} x^{\frac{3}{2}}

= [\frac{1}{3} (2 x^{\frac{3}{2}} ln (x^{2}) - 3 \cdot \frac{8}{9} x^{\frac{3}{2}})]_{1}^{e}

Simplify

[\frac{1}{3} (2 x^{\frac{3}{2}} ln (x^{2}) - 3 \cdot \frac{8}{9} x^{\frac{3}{2}})]_{1}^{e} : [\frac{1}{3} (4 x^{\frac{3}{2}} ln (x) - \frac{8}{3} x^{\frac{3}{2}})]_{1}^{e}

= [\frac{1}{3} (4 x^{\frac{3}{2}} ln (x) - \frac{8}{3} x^{\frac{3}{2}})]_{1}^{e}

Compute the boundaries:

\frac{4 e ^{\frac{3}{2}}}{9} + \frac{8}{9}

= \frac{4 e ^{\frac{3}{2}}}{9} + \frac{8}{9}

Simplify

= \frac{4 e ^{\frac{3}{2}} + 8}{9}

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

\int_{0}^{1.5} e^{- x} cos (x) d x

\int_{- 2}^{2} x^{2} (4 - x^{2}) d x

\int_{0}^{\frac{1}{5}} x ln (5 x) d x

\int_{1}^{2^{\frac{1}{n}}} x^{n - 1} d x

What is the integral from 1 to e of sqrt(x)ln(x^2) ?
The integral from 1 to e of sqrt(x)ln(x^2) is (4e^{3/2}+8)/9