Download App

7.1 inderfinite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.1 inderfinite integral1 Mark
MCQ
$\int_{}^{} {\frac{{{x^2} + 1}}{{{x^4} + 1}}dx = } $
  • A
    $\frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\frac{{{x^2} - 1}}{{2x}}} \right) + c$
  • B
    $\frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\frac{{{x^2} - 1}}{{\sqrt {2x} }}} \right) + c$
  • C
    $\frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\frac{{{x^2} - 1}}{{2\sqrt x }}} \right) + c$
  • ✓
    $\frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\frac{{{x^2} - 1}}{{\sqrt 2 x}}} \right) + c$
✓

Answer

Correct option: D.
$\frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\frac{{{x^2} - 1}}{{\sqrt 2 x}}} \right) + c$
d
(d)$\int_{}^{} {\frac{{{x^2} + 1}}{{{x^4} + 1}}} \,dx = \int_{}^{} {\frac{{\left( {1 + \frac{1}{{{x^2}}}} \right)}}{{\left( {{x^2} + \frac{1}{{{x^2}}}} \right)}}} \,dx = \int_{}^{} {\frac{{\left( {1 + \frac{1}{{{x^2}}}} \right)\,dx}}{{{{\left( {x - \frac{1}{x}} \right)}^2} + 2}}} $
Put $x - \frac{1}{x} = t \Rightarrow \left( {1 + \frac{1}{{{x^2}}}} \right)\,dx = dt,$ then the required integral is $\frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\frac{{{x^2} - 1}}{{\sqrt 2 x}}} \right) + c$.

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

Let $g$ is the inverse function of $f \,\&\, f ' (x) = \frac{{{x^{10}}}}{{\left( {1\,\, + \,\,{x^2}} \right)}}$. If $g(2) = a$ then $g ' (2)$ is equal to
→
For each real number $x$ such that $ - 1 < x < 1,$let $A(x)$ be the matrix ${(1 - x)^{ - 1}}\left[ {\begin{array}{*{20}{c}}1&{ - x}\\{ - x}&1\end{array}} \right]$ and $z = \frac{{x + y}}{{1 + xy}}$ Then
→
Choose the correct answer from the given four options.

For any two matrices A and B, we have:

  1. $\text{AB}=\text{BA}$

  2. $\text{AB}\neq\text{BA}$

  3. $\text{AB}=\text{O}$

  4. None of the above.

→
Let g(x) = 1 + x - [x] and $\text{f(x)}=\begin{cases}-1,&\text{x}<0\\0,&\text{x}=0\\1,&\text{x}>0\end{cases}$ where [x] denotes the greatest integer less than or equal to x. Then for all x, f(g(x)) is equal to:
  1. x
  2. 1
  3. f(x)
  4. g(x)
→
The probability of a bomb hitting a bridge is $\frac{1}{2}$ and two direct hits are needed to destroy it. The least number of bombs required so that the probability of the bridge beeing destroyed is greater then $0.9$, is
→
If $\cos^{-1}\frac{\text{x}}{3}+\cos^{-1}\frac{\text{y}}{2}=\frac{\theta}{2},$ then, $4\text{x}^2-12\text{xy}\cos^2\frac{\theta}{2}+9\text{y}^2=$
  1. $36$
  2. $36-36\cos\theta$
  3. $18-18\cos\theta$
  4. $18+18\cos\theta$
→
Let $E ^{ C }$ denote the complement of an event $E$. Let $E _{1}, E _{2}$ and $E _{3}$ be any pairwise independent events with $P \left( E _{1}\right) > 0$ and $P \left( E _{1} \cap E _{2} \cap E _{3}\right)=0$ Then $P \left( E _{2}^{ C } \cap E _{3}^{ C } / E _{1}\right)$ is equal to
→
A purse contains $4$ copper coins and $3$ silver coins. A second purse contains $6$ copper coins and $4$ silver coins. A purse is chosen randomly and a coin is taken out of it. What is the probability that it is a copper coin?
→
The value of the determinant $\left| {\,\begin{array}{*{20}{c}}4&{ - 6}&1\\{ - 1}&{ - 1}&1\\{ - 4}&{11}&{ - 1\,}\end{array}} \right|$is
→
Three faces of aj ordinary dice are yellow, two faces are red and one face is blue. The dice is rolled 3 times. The probability that yellow red and blue face appear in the first second and third throws respectively, is
→