Question

Evaluate $\int x^2\log xdx$.

• A. $\frac{x^2}{2}\log x-\frac{1}{9}{x}^2+c$
• B. $\frac{x^3}{3}\log x-\frac{1}{9}{x}^2+c$
• C. $\frac{x^3}{3}\log x-\frac{1}{9}{x}^3+c$
• D. $\frac{x^3}{3}\log x+\frac{1}{9}{x}^3+c$

Answer 1

The correct option is C $\frac{x^3}{3}\log x-\frac{1}{9}{x}^3+c$

We have,

$I=\int x^2\log xdx$

We have,

$\int u.vdx=u\int vdx$ $-\int (\frac{du}{dx}\int vdx)dx$

$I=\log x\int x^{2}dx-\int (\frac{d}{dx}\log x\int x^{2}dx)dx$

$I=\log x\frac{x^3}{3}-\int \left(\frac{1}{x}\frac{x^3}{3}\right)dx$

$I=\frac{x^3\log x}{3}-\int \left(\frac{x^2}{3}\right)dx$

$I=\frac{x^3\log x}{3}-\frac{x^3}{9}+C$

Hence, this is the answer.