integration by parts
- Thread starter jeng
- Start date
Hello, jeng!
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?
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?
- Joined
- Apr 12, 2005
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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.
\(\displaystyle \L\int{x^{2}\cdot\ln(x)\,dx\,=\,\int{ln(x)}\,d(\frac{x^{3}}{3})}\,=\,ln(x)\cdot\frac{x^{3}}{3}-\int{\frac{x^{3}}{3}}\,d(ln(x))\,=\,ln(x)\cdot\frac{x^{3}}{3}-\int{\frac{x^{2}}{3}}\,dx\)
Same result, of course.