integration by parts

[jeng]

how do you integrate by parts:

S x^2 lnx dx ?

 

[soroban]

Hello, jeng!

Integrate by parts: \(\displaystyle \L\:\int x^2\cdot\ln x\,dx\)

There are only two choices: $\displaystyle \;$Let $\displaystyle u \,=\, x^2$ . . . or let $\displaystyle u \,=\, \ln x$
Did you try them?

Let: $\displaystyle u\,=\,\ln x\;\;\;\;dv\,=\,x^2\,dx$

Then: $\displaystyle du\,=\,\frac{dx}{x}\;\;\;v\,=\,\frac{1}{3}x^3$

And we have: \(\displaystyle \;\frac{1}{3}x^3\cdot\ln x\,-\,\L\int\)\(\displaystyle \left(\frac{1}{3}x^3\right)\left(\frac{dx}{x}\right) \;= \;\frac{1}{3}x^3\cdot\ln x\,-\,\frac{1}{3}\L\int\)$\displaystyle x^2\,dx$

Can you finish it?

 

[tkhunny]

I just like to demonstrate this alternate notation.

\(\displaystyle \L\int{x^{2}\cdot\ln(x)\,dx\,=\,\int{ln(x)}\,d(\frac{x^{3}}{3})}\,=\,ln(x)\cdot\frac{x^{3}}{3}-\int{\frac{x^{3}}{3}}\,d(ln(x))\,=\,ln(x)\cdot\frac{x^{3}}{3}-\int{\frac{x^{2}}{3}}\,dx\)

Same result, of course.