$\displaystyle \int\ln^2(x)dx$ Hints please?
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Looks like a prime candidate for "Parts". It's almost an eyeball problem. $\displaystyle x\cdot ln^{2}(x) - \int x\cdot 2\cdot ln(x)\cdot \frac{1}{x}\;dx$
soo, $\displaystyle \ln^2(x)-2x\ln(x)+2x+C$ ??
close, try $\displaystyle x\ln^2(x)-2x\ln(x)+2x+C$
So close! Give it another go. This time, don't lose the 'x' on the front.
Originally Posted by iPod soo, $\displaystyle \ln^2(x)-2x\ln(x)+2x+C$ ?? You should get $\displaystyle x\ln^2(x)-2x\ln(x)+2x+C$
woopsy, that was a typo ^^ i meant to right $\displaystyle x\ln^2(x)-2x\ln(x)+2x+C$ but I have seen what you have done tkhunny, using the 'by parts rule, you have integrated it as if it was $\displaystyle \int1.\ln^2(x)$ right?
Originally Posted by iPod $\displaystyle \int1.\ln^2(x)$ right? Yep with $\displaystyle v = \ln^2x $ and $\displaystyle du = 1$
Thanks you lot
...except that I don't use that silly American method. The Russians are much more visual and literal.
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