\[integrate \int\limits_{e}^{e ^{2}}\frac{dx}{x(\ln x)^{2}}\]
\[\int\limits_{e}^{e ^{2}}\frac{dx}{x(\ln x)^{2}}\] Put \[\ln x=t\] so we'll get \[\frac{1}{x} dx =dt\] Can you change the limits?
yup. u = lnx if x = e^2 ; => 2 x = e ; => 1
Great :D so we have now \[\int_{1} ^{2} \frac{dt}{t^2}\] I think you can do now, can't you?
i don't think so. but i know the answer is 1/2 but i don't know the solution. can you please show me?
We know integral of \[\int x^{n} dx=\frac {x^{n+1}}{n+1} where\ n\ne -1\] here we have \[\int_{1}^{2} \frac{dt}{t^2}=\frac{t^{-2+1}}{-2+1}\] on simplification we get \[\int_{1}^{2} \frac{dt}{t^2}=[\frac{-1}{t}]_1^{2}\] Now just apply the limits
Do you understand this?
yup. Thanks a lot @ash2326 . :)
You're welcome:D
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