_ ^ x 1 + x^2 ;dx = - Vidyadip

MCQ

$\int_{}^{} {x\sqrt {1 + {x^2}} } \;dx = $

A
$\frac{{1 + 2{x^2}}}{{\sqrt {1 + {x^2}} }} + c$
B
$\sqrt {1 + {x^2}} + c$
C
$3{(1 + {x^2})^{3/2}} + c$
✓
$\frac{1}{3}{(1 + {x^2})^{3/2}} + c$

✓

Answer

Correct option: D.

$\frac{1}{3}{(1 + {x^2})^{3/2}} + c$

d
(d) Put $1 + {x^2} = t \Rightarrow x\,dx = \frac{{dt}}{2}$ It reduces to $\frac{1}{2}\int_{}^{} {{t^{1/2}}dt} = \frac{1}{2} \times \frac{{{t^{3/2}}}}{{3/2}} = \frac{1}{3}{(1 + {x^2})^{3/2}} + c.$

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

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

$\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)=$

If $f(x) = {e^x}g(x),g(0) = 2,g'(0) = 1$, then $f'(0)$ is

If $\text{x}=\text{t}^2,\text{y}=\text{t}^3$ Then $\frac{\text{d}^2\text{y}}{\text{dx}^2}=$

$\frac{3}{2}$
$\frac{3}{4\text{t}}$
$\frac{3}{2\text{t}}$
$\frac{3\text{t}}{2}$

$V$ is a matrix of order 3 such that $\mid$ adj $V \mid=7$.
Which of these could be $|V|$ ?

$f(x) = \left\{ {\begin{array}{*{20}{c}}
{ - {x^3} + 1\,\,\,\,\,if\,\,\,\,\,\, - \infty < x \leqslant 1} \\
{|x - 1| + \lambda \,\,\,\,if\,\,\,\,\,\,\,\,\,\,\,\,x > 1}
\end{array}} \right.$ then-

Which of the following are true?

Particular solution is a solution of a differential equation containing no arbitrary constants.
Particular Solution is a solution to a differential equation that contains arbitrary, unevaluated constants.
General solution is a solution of a differential equation containing no arbitrary constants.
General Solution is a solution to a differential equation that contains arbitrary, unevaluated constants.

The solution of $\text{x}^{2}+\text{y}^{2}\frac{\text{dy}}{\text{dx}}=4$ is:

x² + y² = 12x + C
x² + y² = 3x + C
x³ + y³ = 3x + C
x³ + y³ = 12x + C

If the events A and B are independent, then $\text{P}(\text{A}\cap\text{B})$ is equal to,

P(A) + P(B)
P(A) - P(B)
P(A) P(B)
$\frac{\text{P(A)}}{\text{P(B)}}$

Four numbers are chosen at random ( without replacement ) from the set $\{1,2,3,..,20\}$

Statement $-1 :$ The probability that the chosen numbers when arranged in some order will form an $A.P.$ is $\frac{1}{{85}}$ .

Statement $-2 :$ If the four chosen numbers form an $A.P.$, then the set of all possible values of common difference is $\left( { \pm 1, \pm 2, \pm 3, \pm 4, \pm 5} \right)$ છે.

The area common to the ellipse $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1$ and $\frac{\text{x}^2}{\text{b}^2}+\frac{\text{y}^2}{\text{a}^2}=1,0<\text{b}<\text{a}$ is:

$(\text{a}+\text{b})^2\tan^{-1}\frac{\text{b}}{\text{a}}$
$(\text{a}+\text{b})^2\tan^{-1}\frac{\text{a}}{\text{b}}$
$4\text{a}+\text{b}\tan^{-1}\frac{\text{b}}{\text{a}}$
$4\text{a}+\text{b}\tan^{-1}\frac{\text{a}}{\text{b}}$