Download App

STD 12 - 7.1 inderfinite integral — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 7.1 inderfinite integral4 Marks
MCQ
$\int_{}^{} {\frac{{dx}}{{{e^x} + {e^{ - x}}}} = } $
  • A
    ${\tan ^{ - 1}}({e^{ - x}})$
  • ✓
    ${\tan ^{ - 1}}({e^x})$
  • C
    $\log ({e^x} - {e^{ - x}})$
  • D
    $\log ({e^x} + {e^{ - x}})$
✓

Answer

Correct option: B.
${\tan ^{ - 1}}({e^x})$
b
(b)$\int_{}^{} {\frac{{dx}}{{{e^x} + {e^{ - x}}}}} = \int_{}^{} {\frac{{{e^x}}}{{{e^{2x}} + 1}}} \,dx = \int_{}^{} {\frac{{dt}}{{{t^2} + 1}}} = {\tan ^{ - 1}}(t)$

$ = {\tan ^{ - 1}}({e^x}) + c$,$\{$Putting ${e^x} = t \Rightarrow {e^x}dx = dt.$$\}$

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

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

For the two positive numbers $a , b$, if $a , b$ and $\frac{1}{18}$ are in a geometric progression, while $\frac{1}{ a }, 10$ and $\frac{1}{ b }$ are in an arithmetic progression, then, $16 a+12 b$ is equal to $.........$.
→
If the line $3x -4y -k = 0 (k > 0)$ touches the circle $x^2 + y^2 -4x -8y -5 = 0$ at $(a, b)$ then $k + a + b$ is equal to :-
→
If $a > 0,$ then $\sqrt {a + \sqrt {a + \sqrt {a + ....\infty } } } $ is
→
Let $A=\{1,3,4,6,9\}$ and $B=\{2,4,5,8,10\}$. Let $R$ be a relation defined on $A \times B$ such that $R =$ $\left\{\left(\left(a_1, b_1\right),\left(a_2, b_2\right)\right): a_1 \leq b_2\right.$ and $\left.b_1 \leq a_2\right\}$. Then the number of elements in the set $R$ is
→
$A = \{1, 2, 3\}$ and $B = \{3, 8\}$, then $(A \cup B) × (A \cap B)$ is
→
The function $f:R \to R$ defined by $f(x) = (x - 1)$ $(x - 2)(x - 3)$ is
→
The number of integral values of $\lambda $ for which $x^2 + y^2 + \lambda x + (1 - \lambda )y + 5 = 0$ is the equation of a circle whose radius cannot exceed $5$ , is
→
If $\theta $ lies in the second quadrant, then the value of $\sqrt {\left( {\frac{{1 - \sin \theta }}{{1 + \sin \theta }}} \right)} + \sqrt {\left( {\frac{{1 + \sin \theta }}{{1 - \sin \theta }}} \right)} $
→
In a triangle $\mathrm{ABC}$, if $|\overrightarrow{\mathrm{BC}}|=3,|\overrightarrow{\mathrm{C}}|=5$ and $|\overrightarrow{\mathrm{BA}}|=7$, then the projection of the vector $\overline{\mathrm{BA}}$ on $\overline{\mathrm{BC}}$ is equal to:
→
Water is drained from a vertical cylindrical tank by opening a valve at the base of the tank. It is known that the rate at which the water level drops is proportional to the square root of water depth $y$, where the constant of proportionality $k$ $>$ $0$ depends on the acceleration due to gravity and the geometry of the hole. If t is measured in minutes and $k$ = $\frac{1}{15}$ then the time to drain the tank if the water is $4$ meter deep to start with is .......... $\min.$
→