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

The integral from 1 to e of (1+x^2)ln(x) is (2e^3+10)/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 (1+x^2)ln(x)

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

\frac{2 e ^{3} + 10}{9}

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{1}{3} (3 + x^{2}) d x = x + \frac{x ^{3}}{9}

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

Simplify

[ln (x) (x + \frac{x ^{3}}{3}) - (x + \frac{x ^{3}}{9})]_{1}^{e} : [\frac{1}{9} (9 x ln (x) + 3 x^{3} ln (x) - 9 x - x^{3})]_{1}^{e}

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

Compute the boundaries:

\frac{2 e ^{3}}{9} + \frac{10}{9}

= \frac{2 e ^{3}}{9} + \frac{10}{9}

Simplify

= \frac{2 e ^{3} + 10}{9}

\int_{1}^{3} 1 - 4 x - x^{2} - 3 d x

\int_{0}^{1} 12 x e^{x} d x

\int_{0}^{\frac{π}{4}} \frac{5}{cos ( x )} d x

\int_{1}^{7} \frac{15}{x ^{2} + 5 x + 4} d x

\int_{1}^{2} - \frac{4}{x ^{2}} d x

What is the integral from 1 to e of (1+x^2)ln(x) ?
The integral from 1 to e of (1+x^2)ln(x) is (2e^3+10)/9