MCQ
Evaluate the integral :$\int {\frac{{\ln \,(6{x^2})}}{x}\,dx} $
  • A
    $\frac{1}{8}{[\ln (6{x^2})]^3}+ C$
  • $\frac{1}{4}[{\ln ^2}(6{x^2})]+ C$
  • C
    $\frac{1}{2}[\ln (6{x^2})]+ C$
  • D
    $\frac{1}{{16}}{[\ln (6{x^2})]^4}+ C$

Answer

Correct option: B.
$\frac{1}{4}[{\ln ^2}(6{x^2})]+ C$
b
$ln(6x^2) = t$

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

Similar questions

If $A = \left( {\begin{array}{*{20}{c}}
2&{ - 1}\\
{ - 7}&4
\end{array}} \right)$ and $B = \left( {\begin{array}{*{20}{c}}
4&1\\
7&2
\end{array}} \right)$ then which of the following is correct
$\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{n}{{n + y}}} \right)^n}$ equals
If the line $l_1: 3 y -2 x =3$ is the angular bisector of the lines $l_2: x - y +1=0$ and $l_3: \alpha x +\beta y +17=0$, then $\alpha^2+\beta^2-\alpha-\beta$ is equal to
In how many ways can five examination papers be arranged so that physics and chemistry papers never come together
The value of $\left| {\,\begin{array}{*{20}{c}}1&{\cos (\beta - \alpha )}&{\cos (\gamma - \alpha )}\\{\cos (\alpha - \beta )}&1&{\cos (\gamma - \beta )}\\{\cos (\alpha - \gamma )}&{\cos (\beta - \gamma )}&1\end{array}} \right|$ is
$3\,\left[ {{{\sin }^4}\,\left( {\frac{{3\pi }}{2} - \alpha } \right) + {{\sin }^4}\,(3\pi + \alpha )} \right]$ $ - 2\,\left[ {{{\sin }^6}\,\left( {\frac{\pi }{2} + \alpha } \right) + {{\sin }^6}(5\pi - \alpha )} \right] = $
The sum of the first $n$ terms of the series $\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{{15}}{{16}} + .........$ is
Water is drained from a vertical cylindrical tank by opening a valve at the base of the tank. It is known that the rate at which the water level drops is proportional to the square root of water depth $y,$ where the constant of proportionality $k > 0$ depends on the acceleration due to gravity and the geometry of the hole. If t is measured in minutes and $k = \frac{1}{{15}}$ then the time to drain the tank if the water is $4$ meter deep to start with is ......... $\min.$
The distance of the point $B\,(i + 2j + 3k)$ from the line which is passing through $A\,(4i + 2j + 2k)$ and which is parallel to the vector $\overrightarrow C = 2i + 3j + 6k$ is
Let the sum of two positive integers be $24.$ If the probability, that their product is not less than $\frac{3}{4}$ times their greatest positive product, is $\frac{ m }{ n }$, where $gcd(m, n) = 1,$ then $n – m$ equals :