Let #160;u=x+ex; #160;v=1+logxThen, #160;u #39;=1+ex; #160;v #39;=1xUsing #160;the #160;quotient #160;rule:ddxuv=vu #39; uv #39;v2ddxx+ex1 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Logarithmic Inequalities

x+ex 1+log x

Question

$\frac{x + e^{x}}{1 + \log x}$

Open in App

Solution

$Let u = x + e^{x}; v = 1 + \log x Then, u' = 1 + e^{x}; v' = \frac{1}{x} Using the quotient rule: \frac{d}{d x} (\frac{u}{v}) = \frac{v u' - u v'}{v^{2}} \frac{d}{d x} (\frac{x + e^{x}}{1 + \log x}) = \frac{(1 + \log x) (1 + e^{x}) - (x + e^{x}) (\frac{1}{x})}{(1 + \log x)^{2}} = \frac{x + x e^{x} + x \log x + x \log x e^{x} - x - e^{x}}{x (1 + \log x)^{2}} = \frac{x \log x + x \log x e^{x} - e^{x} + x e^{x}}{x (1 + \log x)^{2}} = \frac{x \log x (1 + e^{x}) - e^{x} (1 - x)}{x (1 + \log x)^{2}}$

Suggest Corrections

Similar questions

\int e^{x} {(\frac{1 - x}{1 + x^{2}})}^{2} d x is equal to (a) \frac{e^{x}}{1 + x} + C (b) - \frac{e^{x}}{1 + x^{2}} + C (c) \frac{e^{x}}{(1 + x}

I

\int e^{x} (1 - cot x + {cot}^{2} x) d x = - e^{x} cot x + c

I I : \int e^{x} (\frac{1 + x log x}{x}) d x = e^{x} log x + c

\int \frac{e^{x} (1 + x)}{s i n^{2} (x e^{x})} d x =

\int \frac{e^{x} (1 + x)}{{cos}^{2} (x e^{x})} d x

\int

Join BYJU'S Learning Program