MCQ
$\int_{}^{} {\frac{1}{{{{\log }_x}e}}dx = } $
  • A
    $\log {\log _x}e + c$
  • B
    $\frac{1}{{{{({{\log }_x}e)}^2}}} + c$
  • $x\log \left( {\frac{x}{e}} \right) + c$
  • D
    None of these

Answer

Correct option: C.
$x\log \left( {\frac{x}{e}} \right) + c$
c
(c)$\int_{}^{} {\frac{1}{{{{\log }_x}e}}\,dx = \int_{}^{} {{{\log }_e}x\,dx} = x\log x - x + c} $
$ = x({\log _e}x - {\log _e}e) + c = x{\log _e}\left( {\frac{x}{e}} \right) + c.$

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

Similar questions

If $A =$ $\left[ {\begin{array}{*{20}{c}}0&1&2\\1&2&3\\3&a&1\end{array}} \right]$ ,$A^{-1} =$$\left[ {\begin{array}{*{20}{c}}{1/2}&{ - 1/2}&{1/2}\\{ - 4}&3&c\\{5/2}&{ - 3/2}&{1/2}\end{array}} \right]$, then
Let $\overrightarrow{ a }=\alpha \hat{ i }+2 \hat{ j }-\hat{ k }$ and $\overrightarrow{ b }=-2 \hat{ i }+\alpha \hat{ j }+\hat{ k }$, where $\alpha \in R$. If the area of the parallelogram whose adjacent sides are represented by the vectors $\vec{a}$ and $\vec{b}$ is $\sqrt{15\left(\alpha^{2}+4\right)}$, then the value of $2|\vec{a}|^{2}+(\vec{a} \cdot \vec{b})|\vec{b}|^{2}$ is equal to
The function $f(x)=\tan x-x$
If the vectors $3\,i + 2\,j - k$ and $6\,i - 4xj + yk$ are parallel, then the value of $ x$  and  $y $ will be
In a group of $400$ people, $160$ are smokers and nonvegetarian; $100$ are smokers and vegetarian and the remaining $140$ are non-smokers and vegetarian. Their chances of getting a particular chest disorder are $35\, \%, 20 \,\%$ and $10 \,\%$ respectively. A person is chosen from the group at random and is found to be suffering from the chest disorder. The probability that the selected person is a smoker and non-vegetarian is ...... .
Let R be a relation on the set N given by R = {(a, b): a = b - 2, b > 6}. Then,
  1. (2, 4) ∈ R
  2. (3, 8) ∈ R
  3. (6, 8) ∈ R
  4. (8, 7) ∈ R
The value of a so that the sum of the squares of the roots of the equation ${x^2} - (a - 2)x - a + 1 = 0$ assume the least value, is
If 4a + 2b + c = 0, then the equation 3ax2 + 2bx + c = 0 has atleast one real root lying in the interval:
  1. (0, 1)
  2. (1, 2)
  3. (0, 2)
  4. None of these.
The point on the curve y2 = 4x which is nearest to, the point (2, 1) is:
  1. $1,2\sqrt{2}$
  2. $(1, 2)$
  3. $(1, -2)$
  4. $(-2, 1)$
An orthogonal matrix is