_ ^ e^x (1 + x) x ;dx = - Vidyadip

MCQ

$\int_{}^{} {{e^x}(1 + \tan x)\sec x\;dx = } $

A
${e^x}\cot x$
B
${e^x}\tan x$
✓
${e^x}\sec x$
D
${e^x}\cos x$

✓

Answer

Correct option: C.

${e^x}\sec x$

c
(c)$\int_{}^{} {{e^x}(1 + \tan x)\sec x\,dx} = \int_{}^{} {{e^x}\sec x\,dx} + \int_{}^{} {{e^x}\tan x\sec x\,dx} $
$ = {e^x}\sec x - \int_{}^{} {{e^x}\sec x\tan x\,dx} + \int_{}^{} {{e^x}\sec x\tan x\,dx} $
$ = {e^x}\sec x + c.$
Aliter : $\int_{}^{} {{e^x}(\sec x + \sec x\tan x)\,dx} = {e^x}\sec x + c$
Obviously, it is of the form $\int_{}^{} {{e^x}\left\{ {f(x) + f'(x)} \right\}} \,dx.$

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

$\int {\frac{{\cos 2x - 1\,\,}}{{\cos 2x + 1}}dx = } $

For $\alpha \in N$, consider a relation $R$ on $N$ given by $R =\{( x , y ): 3 x +\alpha y$ is a multiple of 7$\}$.The relation $R$ is an equivalence relation if and only if.

The area of the region bounded by the parabola y = x² + 1 and the staight line x + y = 3 is given by:

$\frac{45}{7}$
$\frac{25}{4}$
$\frac{\pi}{18}$
$\frac{9}{2}$

Find the principal values of: $\cot ^{-1}(1)$

The shortest distance between the lines $\frac{\text{x}-3}{3}=\frac{\text{y}-8}{-1}=\frac{\text{z}-3}{1}$ and, $\frac{\text{x}+3}{-3}=\frac{\text{y}+7}{2}=\frac{\text{z}-6}{4}$ is:

If the set A contains 5 elements and the set B contains 6 elements, then the number of one-one and onto mappings from A to B is:

720
120
0
None of these.

The area (sq. units) bounded by the parabola y2 = 4ax and the line x = a and x = 4a is:

$\frac{\text{35a}^2}{3}$
$\frac{4\text{a}^2}{3}$
$\frac{7\text{a}^2}{3}$
$\frac{56\text{a}^2}{3}$

The area of the ellipse $\frac{\text{x}2}{9}+\frac{\text{y}^2}{4}=1$ in first quadrant is $6\pi$ sq. units.
The ellipse is rotated about its centre in anti-clockwise direction till its major axis coincides with y-axis. Now the area of the ellipse in first Quadrant is $\pi$ sq. units.

2
4
6
8

Let $\alpha$ (a) and $\beta$ (a) be the roots of the equation $(\sqrt[3]{1+a}-1) x^2+(\sqrt{1+a}-1) x+(\sqrt[6]{1+a}-1)=0$ where $a>-1$. Then $\lim _{a \rightarrow 0^{+}} \alpha(a)$ and $\lim _{a \rightarrow 0^{+}} \beta(a)$ are

If $x = t + {1 \over t},y = t - {1 \over t},$ then ${{{d^2}y} \over {d{x^2}}}$ is equal to