MCQ
$\int_{\,1}^{\,x} {\frac{{\log {x^2}}}{x}\,dx = } $
- ✓${(\log x)^2}$
- B$\frac{1}{2}{(\log x)^2}$
- C$\frac{{\log {x^2}}}{2}$
- DNone of these
Let $\log x = t$
==> $\frac{{dx}}{x} = dt$
$\therefore I = 2\int_0^{\log x} {t\,dt = 2\,\left[ {\frac{{{t^2}}}{2}} \right]} _0^{\log x} = {(\log x)^2}$.
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$f(x)=\left\{\begin{array}{cc}e^{\min \left[x^2, x-[x]\right\}}, & x \in[0,1) \\e^{\left[x-\log _e x\right]}, & x \in[1,2]\end{array}\right.$
where [t] denotes the greatest integer less than or equal to $t$. Then the value of the integral $\int \limits_0^2 x f(x) d x$ is