What is the integral of (x+1)^2ln(3x) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x+1)^2ln(3x)

\int (x + 1)^{2} ln (3 x) d x

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

Solution steps

\int (x + 1)^{2} ln (3 x) d x

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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

\int \frac{1}{u ^{4} + u} d u

\int \frac{x ^{3} + x - 4}{x - 1} d x

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

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