_ ^ 1 ( x^2 + a^2 )( x^2 + b^2 ) dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{1}{{({x^2} + {a^2})({x^2} + {b^2})}}dx = } $

✓
$\frac{1}{{({a^2} - {b^2})}}\left[ {\frac{1}{b}{{\tan }^{ - 1}}\left( {\frac{x}{b}} \right) - \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{x}{a}} \right)} \right] + c$
B
$\frac{1}{{({b^2} - {a^2})}}\left[ {\frac{1}{b}{{\tan }^{ - 1}}\left( {\frac{x}{b}} \right) - \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{x}{a}} \right)} \right] + c$
C
$\frac{1}{b}{\tan ^{ - 1}}\left( {\frac{x}{b}} \right) - \frac{1}{a}{\tan ^{ - 1}}\left( {\frac{x}{a}} \right) + c$
D
$\frac{1}{a}{\tan ^{ - 1}}\left( {\frac{x}{a}} \right) - \frac{1}{b}{\tan ^{ - 1}}\left( {\frac{x}{b}} \right) + c$

✓

Answer

Correct option: A.

$\frac{1}{{({a^2} - {b^2})}}\left[ {\frac{1}{b}{{\tan }^{ - 1}}\left( {\frac{x}{b}} \right) - \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{x}{a}} \right)} \right] + c$

a
(a)$\int_{}^{} {\frac{1}{{({x^2} + {b^2})({x^2} + {a^2})}}} \,dx$
$ = \frac{1}{{{a^2} - {b^2}}}\int_{}^{} {\left[ {\frac{1}{{{x^2} + {b^2}}} - \frac{1}{{{x^2} + {a^2}}}} \right]} \,dx$
$ = \frac{1}{{({a^2} - {b^2})}}\left[ {\frac{1}{b}{{\tan }^{ - 1}}\left( {\frac{x}{b}} \right) - \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{x}{a}} \right)} \right] + c$.
Note : Students should remember this question as a formula.

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 - 7.1 inderfinite integral→STD 12 - 7.1 inderfinite integral — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

For the function $\text{f}(\text{x})=\text{x}+1\text{x},\text{x}\in[1,3],$ the value of $c$ for the Lagrange's mean value theorem is :

The sum $\sum\limits_{n = 1}^\infty {{{\cot }^{ - 1}}} \left( {\frac{{2\left( {\sum\limits_{k = 1}^n k } \right) - 1}}{3}} \right)$ is equal to

If $a = i - 2j + 3k$ and $b = 3i + j + 2k,$ then the unit vector perpendicular to $a$ and $b$ is

Which of the following is the integrating factor of $(\text{x}\log\text{x})\frac{\text{dy}}{\text{dx}}+\text{y}=2\log\text{x}?$

The ratio in which the $x-$ axis divides the area of the region bounded by the curves $y = x^2 - 4 x \,\, \& \,\, y = 2 x - x^2$ is :

Let $\text{f}:\text{R}-\Big\{-\frac{4}{3}\Big\}\rightarrow\text{R}$ be a function defined as $f(\text{x})=\frac{4\text{x}}{3\text{x}+4}.$ The inverse of f is the map g: Range $\text{f}:\text{R}-\Big\{-\frac{4}{3}\Big\}\rightarrow\text{R}$ given by:

$A = \left[ {\begin{array}{*{20}{c}}
{{{10}^{30}} + 5}&{{{10}^{20}} + 4}&{{{10}^{20}} + 6}\\
{{{10}^4} + 2}&{{{10}^8} + 7}&{{{10}^{10}} + 2n}\\
{{{10}^4} + 8}&{{{10}^6} + 4}&{{{10}^{15}} + 9}
\end{array}} \right]$ ,

$n \in N$, then

The value of a for which the function $\text{f(x)}=\begin{cases}\frac{(4^\text{x}-1)^3}{\sin\Big(\frac{\text{x}}{\text{a}}\Big)\log\Big\{\Big(1+\frac{\text{x}^2}{3}\Big)\Big\}},&\text{x}\neq0\\12(\log4)^3,&\text{x}=0\end{cases}$ may be continuous at x = 0 is:

If $\text{x},\text{ y}\in\text{R},$ then the determinant $\triangle=\begin{vmatrix}\cos\text{x}&-\sin\text{x}&1\\\sin\text{x}&\cos\text{x}&1\\\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&0\end{vmatrix}$ lies in the interval:

$\int_{}^{} {\frac{{\sin 2x}}{{{{\sin }^4}x + {{\cos }^4}x}}dx = } $