_ ^ 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

In which of these intervals is the function $f(x)=3 x^2-4 x$ strictly decreasing?

$\int_{}^{} {\frac{{{e^{5\log x}} - {e^{4\log x}}}}{{{e^{3\log x}} - {e^{2\log x}}}}\;dx = } $

If $\tan^{-1}\frac{\text{x}+1}{\text{x}-1}+\tan^{-1}\frac{\text{x}-1}{\text{x}}=\tan^{-1}(-7),$ then the value of x is:

Area of the region bounded by the curve $\text{y}=\cos\text{x}$ between $x = 0$ and $\text{x}=\pi$ is:

Area bounded by curve $xy = c,$ $x - $ axis between $x = 1$ and $x = 4,$ is

If the vectors $ai + j + k,\,\,i + bj + k$ and $i + j + ck$ $(a \ne b \ne c \ne 1)$ are coplanar, then the value of $\frac{1}{{1 - a}} + \frac{1}{{1 - b}} + \frac{1}{{1 - c}} = $

Let $\int_\alpha^{\log _e^4} \frac{\mathrm{dx}}{\sqrt{\mathrm{e}^{\mathrm{x}}-1}}=\frac{\pi}{6}$. Then $\mathrm{e}^\alpha$ and $\mathrm{e}^{-\alpha}$ are the roots of the equation :

The least value of the product $xyz$ for which the determinant $\left| {\begin{array}{*{20}{c}}
x&1&1 \\
1&y&1 \\
1&1&z
\end{array}} \right|$ is non-negative, is

Which of the following transformation reduce the differential quation into the form $\frac{\text{du}}{\text{dx}}+\text{P}(\text{x})\text{u}=\text{Q}(\text{x})$ into the from $\frac{\text{dz}}{\text{dx}}+\frac{\text{z}}{\text{x}}\log\text{z}=\frac{\text{z}}{\text{x}^{2}}(\log\text{z})^{2}$

If $\overrightarrow{ a } \cdot \overrightarrow{ b }=1, \overrightarrow{ b } \cdot \overrightarrow{ c }=2$ and $\overrightarrow{ c } \cdot \overrightarrow{ a }=3$, then the value of $[\vec{a} \times(\vec{b} \times \vec{c}), \vec{b} \times(\vec{c} \times \vec{a}), \vec{c} \times(\vec{b} \times \vec{a})]$ is