What is the integral of (x^2-x+1)ln(x) ?

The integral of (x^2-x+1)ln(x) is 1/3 x^3ln(x)-1/2 x^2ln(x)+xln(x)-(x^3)/9+(x^2)/4-x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (x^2-x+1)ln(x)

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

\frac{1}{3} x^{3} ln (x) - \frac{1}{2} x^{2} ln (x) + x ln (x) - \frac{x ^{3}}{9} + \frac{x ^{2}}{4} - x + C

Solution steps

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

Apply Integration By Parts

= ln (x) (\frac{x ^{3}}{3} - \frac{x ^{2}}{2} + x) - \int \frac{1}{6} (2 x^{2} - 3 x + 6) d x

\int \frac{1}{6} (2 x^{2} - 3 x + 6) d x = \frac{x ^{3}}{9} - \frac{x ^{2}}{4} + x

= ln (x) (\frac{x ^{3}}{3} - \frac{x ^{2}}{2} + x) - (\frac{x ^{3}}{9} - \frac{x ^{2}}{4} + x)

Simplify

ln (x) (\frac{x ^{3}}{3} - \frac{x ^{2}}{2} + x) - (\frac{x ^{3}}{9} - \frac{x ^{2}}{4} + x) : \frac{1}{3} x^{3} ln (x) - \frac{1}{2} x^{2} ln (x) + x ln (x) - \frac{x ^{3}}{9} + \frac{x ^{2}}{4} - x

= \frac{1}{3} x^{3} ln (x) - \frac{1}{2} x^{2} ln (x) + x ln (x) - \frac{x ^{3}}{9} + \frac{x ^{2}}{4} - x

Add a constant to the solution

= \frac{1}{3} x^{3} ln (x) - \frac{1}{2} x^{2} ln (x) + x ln (x) - \frac{x ^{3}}{9} + \frac{x ^{2}}{4} - x + C

What is the integral of (x^2-x+1)ln(x) ?
The integral of (x^2-x+1)ln(x) is 1/3 x^3ln(x)-1/2 x^2ln(x)+xln(x)-(x^3)/9+(x^2)/4-x+C