Download App

7.1 inderfinite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.1 inderfinite integral1 Mark
MCQ
$\int_{}^{} {\frac{1}{{{x^3}}}{{[\log {x^x}]}^2}\;dx = } $
  • A
    $\frac{{{x^3}}}{3}(\log x) + x + c$
  • ✓
    $\frac{1}{3}{(\log x)^3} + c$
  • C
    $3\log (\log x) + c$
  • D
    None of these
✓

Answer

Correct option: B.
$\frac{1}{3}{(\log x)^3} + c$
b
(b)$\int_{}^{} {\frac{1}{{{x^3}}}{{[\log {x^x}]}^2}\,dx} = \int_{}^{} {\frac{1}{{{x^3}}}{{[x\log x]}^2}dx} $
$ = \int_{}^{} {\frac{1}{x}{{(\log x)}^2}dx} = \frac{1}{3}{(\log x)^3} + c$, $\{$Putting $\log x = t\} $.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from 7.1 inderfinite integral→7.1 inderfinite integral — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Consider set $A = \{1,2,3\}$ . Number of symmetric relations that can be defined on $A$ containing the ordered pair $(1,2)$ & $(2,1)$ is
→
Direction cosines of the line $\frac{x-1}{2}=\frac{1-y}{3}=\frac{2 z-1}{12}$ are:
→
Let the area of the region enclosed by the curves $y=3 x, 2 y=27-3 x$ and $y=3 x-x \sqrt{x}$ be $A$. Then $10 \mathrm{~A}$ is equal to
→
Find the principal value of $\tan ^{-1}(-\sqrt{3})$
→
Let $f(x)$ be a non-negative continous function such that the area bounded by the curve $y = f(x)$, $x -$ axis and the ordinates $x = \frac{\pi }{4}$, $x = \beta > \frac{\pi }{4}$ is $\left( {\beta \sin \beta + \frac{\pi }{4}\cos \beta + \sqrt 2 \beta } \right)$. Then $f\;\left( {\frac{\pi }{2}} \right)$ is
→
If $x=\sin y$, then $\frac{d^2 y}{d x^2}=$ __________ . , $(0< x <1)$.
→
If $\int \limits_{\frac{1}{3}}^3\left|\log _e x\right| d x=\frac{m}{n} \log _e\left(\frac{n^2}{e}\right)$, where $m$ and $n$ are coprime natural numbers, then $m ^2+ n ^2-5$ is equal to $............$.
→
$\int_{}^{} {\frac{{1 + {{\cos }^2}x}}{{{{\sin }^2}x}}dx} = $
→
If $\text{A}=\begin{bmatrix} \text{a} & 0 & 0 \\ 0 & \text{a} & 0 \\ 0 & 0 &\text{a} \end{bmatrix},$ then the value of |adj A| is:
  1. a27
  2. a9
  3. a6
  4. a2
→
${d \over {dx}}\left( {{{\tan }^{ - 1}}\sqrt {{{1 + \cos {x \over 2}} \over {1 - \cos {x \over 2}}}} } \right)$ is equal to
→