int_{1}^{e} frac{dx}{x(1+log x)^{2}} = | vedclass

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(A) माना $I = \int_{1}^{e} \frac{dx}{x(1+\log x)^{2}}$.$u = 1 + \log x$ प्रतिस्थापित करने पर,$du = \frac{1}{x} dx$ प्राप्त होता है।जब $x = 1$,तब $u =

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$\frac{1}{2}$

Vedclass · Answer

(A) माना $I = \int_{1}^{e} \frac{dx}{x(1+\log x)^{2}}$.$u = 1 + \log x$ प्रतिस्थापित करने पर,$du = \frac{1}{x} dx$ प्राप्त होता है।जब $x = 1$,तब $u = 1 + \log 1 = 1 + 0 = 1$.जब $x = e$,तब $u = 1 + \log e = 1 + 1 = 2$.अतः,$I = \int_{1}^{2} \frac{du}{u^{2}} = \int_{1}^{2} u^{-2} du$.$I = \left[ \frac{u^{-1}}{-1} \right]_{1}^{2} = \left[ -\frac{1}{u} \right]_{1}^{2}$.$I = -\left( \frac{1}{2} - \frac{1}{1} \right) = -\left( -\frac{1}{2} \right) = \frac{1}{2}$.

$\int_{1}^{e} \frac{dx}{x(1+\log x)^{2}} =$

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