MCQ
$\int e^{5 \log x} d x$ is equal to
  • A
    $\frac{x^5}{5}+C$
  • $\frac{x^6}{6}+C$
  • C
    $5 x^4+C$
  • D
    $6 x^5+C$

Answer

Correct option: B.
$\frac{x^6}{6}+C$
$\text {Let } I=\int e^{5 \log x} d x$
$=\int e^{\log x^5} d x=\int x^5 d x $
$\left[\because e^{\log x}=x\right]$
$=\frac{x^6}{6}+C$

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

Let $f(x)$ be a positive function such that the area bounded by $y=f(x), y=0$ from $x=0$ to $x=a>0$ is $\mathrm{e}^{-\mathrm{a}}+4 \mathrm{a}^2+\mathrm{a}-1$. Then the differential equation, whose general solution is $y=c_1 f(x)+c_2$, where $c_1$ and $c_2$ are arbitrary constants, is :
If $A = \left( {\begin{array}{*{20}{c}}i&1\\0&i\end{array}} \right)$, then ${A^4}$ equals
The area enclosed between the curves $y=x|x|$ and $\mathrm{y}=\mathrm{x}-|\mathrm{x}|$ is :
If $\text{f(x)}=\begin{cases}\frac{1-\cos10\text{x}}{\text{x}^2},&\text{x}<0\\\text{a},&\text{x}=0\\\frac{\sqrt{\text{x}}}{\sqrt{625+\sqrt{\text{x}}}-25},&\text{x}>0\end{cases}$ then the value of so that f(x) may be continuous at x = 0 is:
Find the value of x if $ \sin (\text{arc} \sin \text{x}) = \frac {\sqrt {2}}{4}:$
For the function $f (x) =$ $\frac{1}{{x + {2^{\frac{1}{{(x - 2)}}}}}}$ , $x \ne 2$ which of the following holds ?
Let $x(t)=2 \sqrt{2} \cos t \sqrt{\sin 2 t}$ and $y ( t )=2 \sqrt{2} \sin t \sqrt{\sin 2 t }, t \in\left(0, \frac{\pi}{2}\right)$. Then $\frac{1+\left(\frac{ dy }{ dx }\right)^{2}}{\frac{ d ^{2} y }{ dx ^{2}}}$ at $t =\frac{\pi}{4}$ is equal to..
$\int_{\,2}^{\,3} {\frac{{dx}}{{{x^2} - x}} = } $
Let three vectors $\vec{a}, \overrightarrow{\mathrm{b}}$ and $\vec{c}$ be such that $\vec{a} \times \overrightarrow{\mathrm{b}}=\vec{c}, \overrightarrow{\mathrm{b}} \times \vec{c}=\vec{a}$ and $|\vec{a}|=2$

Then which one of the following is not true?

If ${\tan ^{ - 1}}2x + {\tan ^{ - 1}}3x = \frac{\pi }{4}$, then $x =$