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

MCQ

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

✓
$\frac{1}{2}[{\sin ^{ - 1}}{x^2} + \sqrt {1 - {x^4}} ] + c$
B
$\frac{1}{2}[{\sin ^{ - 1}}{x^2} + \sqrt {1 - {x^2}} ] + c$
C
${\sin ^{ - 1}}{x^2} + \sqrt {1 - {x^4}} + c$
D
${\sin ^{ - 1}}{x^2} + \sqrt {1 - {x^2}} + c$

✓

Answer

Correct option: A.

$\frac{1}{2}[{\sin ^{ - 1}}{x^2} + \sqrt {1 - {x^4}} ] + c$

a
(a)$\int_{}^{} {x\sqrt {\frac{{1 - {x^2}}}{{1 + {x^2}}}} } dx = \int_{}^{} {\frac{{x.\,(1 - {x^2})}}{{\sqrt {1 - {x^4}} }}} dx$
$\{$ Multiplying ${N^r}$ and ${D^r}$ by ${(1 - {x^2})^{1/2}}\} $
$ = \int_{}^{} {\frac{x}{{\sqrt {1 - {x^4}} }}} \,dx - \int_{}^{} {\frac{{{x^3}}}{{\sqrt {1 - {x^4}} }}} dx$

$ = \frac{1}{2}[{\sin ^{ - 1}}({x^2}) + \sqrt {1 - {x^4}} ] + c$.
(By putting ${x^2} = t$ and $\sqrt {1 - {x^4}} = \sqrt t $respectively)

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

Let $\vec{a}$ and $\vec{b}$ be two vector such that $|\vec{a}|=\sqrt{14}$, $|\vec{b}|=\sqrt{6}$ and $|\vec{a} \times \vec{b}|=\sqrt{48}$. Then $(\vec{a} \cdot \vec{b})^2$ is equal to $...........$.

The function $\text{f{x}}\begin{cases}1,&|\text{x}|\geq1\\\frac{1}{\text{n}^2},&\frac{1}{\text{n}}<|\text{x}|<\frac{1}{\text{n}-1},\text{n}=2,3,...\end{cases}$

Is discontinuous at finitely many points.
Is continuous everywher.
Is discontinuous only at $\text{x}=\pm\frac{1}{\text{n}},\text{n }\in\text{ z}-\{0\}$ and x = 0
None of these.

Let $\alpha x=\exp \left(x^\beta y^\gamma\right)$ be the solution of the differential equation $2 x^2 y d y-\left(1-x y^2\right) d x=0$, $x >0, y (2)=\sqrt{\log _e 2}$. Then $\alpha+\beta-\gamma$ equals :

Choose the correct answers from the given four options:
If $\text{f(x)}=\text{x}^2\sin\frac{1}{\text{x}},$ where $\text{x}\neq0,$ then the value of the function f at x = 0, so that the function is continuous at x = 0, is:

0
-1
1
None of these

$\int_{}^{} {\frac{{{x^2} + x - 6}}{{(x - 2)(x - 1)}}dx = } $

Let $f: R \rightarrow R$ be defined as

$f(\mathrm{x})= -\frac{4}{3} x^{3}+2 x^{2}+3 x ,\quad x>0$

$\quad\quad\quad\quad 3 x e^{x}, \quad\quad\quad\quad\quad\quad\mathrm{x} \leq 0$

Then $\mathrm{f}$ is increasing function in the interval.

If the multiplicative group of $2 × 2$ matrices of the form $\left( {\begin{array}{*{20}{c}}a&a\\a&a\end{array}} \right)$, for $a \ne 0$ and $a \in R$, then the inverse of $\left( {\begin{array}{*{20}{c}}2&2\\2&2\end{array}} \right)$ is

Area of the figure enclosed by $y = {\cos ^{ - 1}}(\cos \,x),x\, \in \,\left[ {2\pi ,4\pi } \right]$ ; $x-$ axis and $y = {\tan ^{ - 1}}\,x + {\tan ^{ - 1}}\,\frac{1}{x}$ is

If $\vec{a}$ is a nonzero vector of magnitude $^{\prime} a^{\prime}$ and $\lambda$ a nonzero scalar, then $\lambda \,\,\vec{a}$ is unit vector if

A linear programming of linear functions deals with:

Minimizing
Optimizing
Maximizing
None