_0^ x^2 ,dx ( x^2 + a^2 )( x^2 + b^2 ) = - Vidyadip

MCQ

$\int_0^\infty {\frac{{{x^2}\,dx}}{{({x^2} + {a^2})({x^2} + {b^2})}}} = $

A
$\frac{\pi }{{2(a - b)}}$
B
$\frac{\pi }{{2(b - a)}}$
C
$\frac{\pi }{{(a + b)}}$
✓
$\frac{\pi }{{2(a + b)}}$

✓

Answer

Correct option: D.

$\frac{\pi }{{2(a + b)}}$

d
(d) $\int_0^\infty {\frac{{{x^2}dx}}{{({x^2} + {a^2})({x^2} + {b^2})}} = \int_0^\infty {\frac{{({x^2} + {a^2}) - {a^2}}}{{({x^2} + {a^2})({x^2} + {b^2})}}{\rm{ }}} dx} $

$ = \int_0^\infty {\frac{1}{{{x^2} + {b^2}}}dx - {a^2}\int_0^\infty {\frac{1}{{({x^2} + {a^2})({x^2} + {b^2})}}{\rm{ }}} dx} $

$ = \left[ {\frac{1}{b}{{\tan }^{ - 1}}\frac{x}{b}} \right]_0^\infty - \frac{{{a^2}}}{{({a^2} - {b^2})}}\int_0^\infty {\left( {\frac{1}{{{x^2} + {b^2}}} - \frac{1}{{{x^2} + {a^2}}}} \right){\rm{ }}} dx$

$ = \frac{1}{b}.\frac{\pi }{2} - \frac{{{a^2}}}{{({a^2} - {b^2})}}\left[ {\frac{1}{b}{{\tan }^{ - 1}}\frac{x}{b} - \frac{1}{a}{{\tan }^{ - 1}}\frac{x}{a}} \right]_0^\infty $

$= \frac{\pi }{{2(a + b)}}$.

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 STD 12 - 7.2 definite integral→STD 12 - 7.2 definite integral — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

Given $\int\limits_0^{\frac{\pi }{2}} {\,\,\frac{{dx}}{{1 + \sin x + \cos x}}} = ln\, 2$, then the value of the def. integral. $\int\limits_0^{\frac{\pi }{2}} {\,\,\frac{{\sin \,x}}{{1 + \sin x + \cos x}}} \,dx$ is equal to

If $a \ne b \ne c,$ the value of $x$ which satisfies the equation $\left| {\,\begin{array}{*{20}{c}}0&{x - a}&{x - b}\\{x + a}&0&{x - c}\\{x + b}&{x + c}&0\end{array}\,} \right| = 0$, is

Find the values of $a, b, c$ and $d$ respectively if $\left[\begin{array}{cc}2 a+b & a-2 b \\ 5 c-d & 4 c+3 d\end{array}\right]=\left[\begin{array}{cc}4 & -3 \\ 11 & 24\end{array}\right]$.

The function $f : R \rightarrow R$ defined by $f(x) = 2^x + 2^{|x|}$ is:

The global maximum value of

$f(x) = log_{10}(4x^3 -12x^2 + 11x -3)$, $x \in \left[ {2,3} \right]$, is

Let $\left| {\,\begin{array}{*{20}{c}}{6i}&{ - 3i}&1\\4&{3i}&{ - 1}\\{20}&3&i\end{array}\,} \right| = x + iy$, then

$\int\frac{\text{x}^3}{\sqrt{1+\text{x}^2}}\text{ dx}=\text{a}(1+\text{x}^2)^{\frac{3}{2}}+\text{b}\sqrt{1+\text{x}^2}+\text{C},$ then :

The function ${x^5} - 5{x^4} + 5{x^3} - 1$ is

Consider $P(1,2,-3)$ , $Q(-2,1,-4)$ , $R(3,4,-2)$ and $\vec B = {A_x}\hat i + {A_y}\hat j + {A_z}\hat k$ .If $A_x, A_y$ and $A_z$ be projections of area of triangle $PQR$ on the $yz, zx$ and $xy$ planes respectively, then value of ${\left| {\vec B} \right|^2}$ is

If the solution curve $y=y(x)$ of the differential equation $\left(1+y^2\right)\left(1+\log _6 x\right) d x+x d y=0, x>0$ passes through the point $(1,1)$ and $y(\mathrm{e})=\frac{\alpha-\tan \left(\frac{3}{2}\right)}{\beta+\tan \left(\frac{3}{2}\right)}$, then $\alpha+2 \beta$ is . . . . . . . . .