Download App

Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
MCQ
Choose the correct answer in Exercises:
$\int\frac{10\text{x}^9+10^{\text{x}}\log_\text{e}10}{\text{x}^{10}+10^{\text{x}}}\text{ equals}$
  • A
    $10^\text{x}-\text{x}^{10}+\text{C}$
  • B
    $10^\text{x}+\text{x}^{10}+\text{C}$
  • C
    $(10^\text{x}-\text{x}^{10})^{-1}+\text{C}$
  • ✓
    $\log(10^\text{x}+\text{x}^{10})+\text{C}$
✓

Answer

Correct option: D.
$\log(10^\text{x}+\text{x}^{10})+\text{C}$
$\text{Let I}=\int\frac{10\text{x}^9+10^{\text{x}}\log_\text{e}10}{\text{x}^{10}+10^{\text{x}}}\text{ dx} \ \ \ \ ...\text{(i)} $
Putting ${\text{x}^{10}+10^{\text{x}}}=\text{t}\ \ \ \ \Rightarrow\ \ \ \ (10\text{x}^9+10^{\text{x}}\log_\text{e}10)\text{ dx = dt} $
$\therefore\ \ \ \ \ $From eq. (i), $\text{I}=\int\frac{\text{dt}}{\text{t}}=\log\begin{vmatrix}\text {t}\end{vmatrix}+\text{c}=\log\begin{vmatrix}\text{x}^{10}+10^\text{x}\end{vmatrix}+\text{c}$
Therefore, option (D) is correct.

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 Integrals→Integrals — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Let $\alpha \in R$ be such that the function

$f(x)=\left\{\begin{array}{ll} \frac{\cos ^{-1}\left(1-\{x\}^{2}\right) \sin ^{-1}(1-\{x\})}{\{x\}-\{x\}^{3}}, & x \neq 0 \\ \alpha, & x=0 \end{array}\right.$

is continuous at $x=0,$ where $\{x\}=x-[x],[x]$ is the greatest integer less than or equal to $X$.

Then :

→
If for a continuous function $f(x),$ $\int\limits_{ - \pi }^t {(f(x) + x\,\,dx)}  = {\pi ^2} - {t^2},$ for all  $t\, \ge  - \pi ,$ then $f\left( { - \frac{\pi }{3}} \right)$ is equal to
→
If $x = a\sin 2\theta (1 + \cos 2\theta ),y = b\cos 2\theta (1 - \cos 2\theta )$, then ${{dy} \over {dx}} = $
→
If $y = {x^{{x^{x......\infty }}}}$, then ${{dy} \over {dx}} = $
→
$3 x^2-6 x+5$ is an increasing function, if
→
If the value of the integral $\int_{0}^{5} \frac{x+[x]}{e^{x-[x]}} \,d x=\alpha e^{-1}+\beta$

where $\alpha, \beta \in R, 5 \alpha+6 \beta=0$, and $[\mathrm{x}]$ denotes the greatest integer less than or equal to $x$; then the value of $(\alpha+\beta)^{2}$ is equal to :

→
Choose the correct answer from the given four options.
The feasible solution for a LPP is shown in. Let Z = 3x - 4y be the objective function.

Minimum of Z occurs at:
→
If $a = 4i + 6j$ and $b = 3\,j + 4\,k,$ then the component of $a $ along $ b $ is
→
The area bounded by the curves $y = x (1 - \ln x) ; \,\,x = e^{-1}$ and positive $X-$ axis between $x = e^{-1}$ and $x = e$ is :
→
If $ab = 2a + 3b,\,a > 0,\,\,b > 0$ then the minimum value of  $ ab$  is
→