MCQ
$\int_{}^{} {\frac{{{e^{{{\tan }^{ - 1}}x}}}}{{1 + {x^2}}}dx = } $
- A$\log (1 + {x^2}) + c$
- B$\log {e^{{{\tan }^{ - 1}}x}} + c$
- ✓${e^{{{\tan }^{ - 1}}x}} + c$
- D${\tan ^{ - 1}}{e^{{{\tan }^{ - 1}}x}} + c$
✓
Answer
Correct option: C.
${e^{{{\tan }^{ - 1}}x}} + c$
c
(c) Putting $t = {\tan ^{ - 1}}x \Rightarrow dt = \frac{1}{{1 + {x^2}}}\,dx,$ we get
$\int_{}^{} {\frac{{{e^{{{\tan }^{ - 1}}x}}}}{{1 + {x^2}}}\,dx} = \int_{}^{} {{e^t}dt} $$ = {e^t} + c = {e^{{{\tan }^{ - 1}}x}} + c.$
(c) Putting $t = {\tan ^{ - 1}}x \Rightarrow dt = \frac{1}{{1 + {x^2}}}\,dx,$ we get
$\int_{}^{} {\frac{{{e^{{{\tan }^{ - 1}}x}}}}{{1 + {x^2}}}\,dx} = \int_{}^{} {{e^t}dt} $$ = {e^t} + c = {e^{{{\tan }^{ - 1}}x}} + c.$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are unit vectors, then which of the following values of $\vec{\text{a}}.\vec{\text{b}}$ is not possible?
→The points $D, E$ and $F$ are the mid-points of $A B, B C$ and $C A$ respectively.
What is the area of the shaded region?
→What is the area of the shaded region?
If $y = x\log \left( {{x \over {a + bx}}} \right)$, then ${x^3}{{{d^2}y} \over {d{x^2}}} = $
→The area bounded by $x^2 + y^2 - 2 x = 0\,\, \&\,\, y = \sin \frac{{\pi \,x}}{2}$ in the upper half of the circle is :
→If $f(x) = \cos (\log x)$, then $f({x^2})f({y^2}) - \frac{1}{2}\left[ {f\,\left( {\frac{{{x^2}}}{2}} \right) + f\left( {\frac{{{x^2}}}{{{y^2}}}} \right)} \right]$ has the value
→A unit vector perpendicular to each of the vector $2i - j + k$ and $3i + 4j - k$ is equal to
→The maximum value of $Z = 3x + 4y$ subjected to contraints $\text{x}+\text{y}\leq40,\text{x}+2\text{y}\leq60,\text{x}\geq0$ and $\text{y}\geq0$ is:
→The area of the ellipse $\frac{\text{x}2}{9}+\frac{\text{y}^2}{4}=1$ in first quadrant is $6\pi$ sq. units.
The ellipse is rotated about its centre in anti $-$ clockwise direction till its major axis coincides with $y-$ axis. Now the area of the ellipse in first Quadrant is $\pi$ sq. units.
→The ellipse is rotated about its centre in anti $-$ clockwise direction till its major axis coincides with $y-$ axis. Now the area of the ellipse in first Quadrant is $\pi$ sq. units.
The __________ is the method available for solving an L.P.P
Choose the correct answers from the given four options:
The derivative of $\cos^{-1}(2\text{x}^2-1)$ w.r.t. $\cos^{-1}\text{x}$ is:
→The derivative of $\cos^{-1}(2\text{x}^2-1)$ w.r.t. $\cos^{-1}\text{x}$ is: