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Question Number 48127 by cesar.marval.larez@gmail.com last updated on 19/Nov/18

Answered by tanmay.chaudhury50@gmail.com last updated on 19/Nov/18

$$30)t=\frac{-1}{x}dt=\frac{1}{x^2}dx$$$$\int e^{t}dt=e^t+c$$$$e^t+c$$$$=e^{\frac{-1}{x}}+c$$$$35)\int \frac{ln5x}{x}dx$$$$t=ln5x\frac{}{}$$$$\frac{dt}{dx}=\frac{1}{5x}\times \frac{d}{dx}\left(5x\right)=\frac{1}{5x}\times 5=\frac{1}{x}dt=\frac{dx}{x}$$$$\int tdt=\frac{t^2}{2}+c$$$$=\frac{{\left(ln5x\right)}^2}{2}+c$$$$38)\int \frac{1}{x{\left(lnx\right)}^2}dx$$$$t=lnx\frac{dt}{dx}=\frac{1}{x}dt=\frac{dx}{x}$$$$\int \frac{dt}{t^2}$$$$\int t^{-2}dt$$$$=\frac{t^{-2+1}}{-2+1}+c$$$$=\frac{-1}{t}+c$$$$=\frac{-1}{lnx}+c$$

Commented by cesar.marval.larez@gmail.com last updated on 19/Nov/18

$$Myfriendthanks.The30iknewbuti$$$$dontfeltsafe$$

Commented by tanmay.chaudhury50@gmail.com last updated on 19/Nov/18

$$mostwelcome...$$

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