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Q. The value of the integral $\displaystyle \int_1^e$ $\frac{1+\log x}{3x} dx$ is equal to

KEAMKEAM 2014Integrals

Solution:

Let $ I =\int\limits_{1}^{e} \frac{1+\log x}{3 x} d x $
$=\int\limits_{1}^{e} \frac{1}{3 x} d x+\frac{1}{3} \int\limits_{1}^{e} \frac{\log x}{x} d x $
Put $\log x=t $
$\Rightarrow \frac{1}{x} d x=d t $
$\therefore I =\int\limits_{1}^{e} \frac{1}{3 x} d x+\frac{1}{3} \int_{0}^{1} t \,d t $
$ =\left[\frac{1}{3} \log x\right]^{e}+\frac{1}{3}\left[\frac{t^{2}}{2}\right]_{0}^{1} $
$ =\frac{1}{3}[\log e-\log 1]+\frac{1}{6}\left[1^{2}-0\right] $
$ =\frac{1}{3}[1-0]+\frac{1}{6}=\frac{3}{6}=\frac{1}{2} $