Download App

7.1 inderfinite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.1 inderfinite integral1 Mark
MCQ
A primitive of $f(x) = \frac{x}{{1 + {x^2}}}$  $dx$=
  • A
    ${\log _e}({x^2} + 1)$
  • B
    $x{\tan ^{ - 1}}x$
  • ✓
    $\frac{{{{\log }_e}({x^2} + 1)}}{2}$
  • D
    $\frac{1}{2}x{\tan ^{ - 1}}x$
✓

Answer

Correct option: C.
$\frac{{{{\log }_e}({x^2} + 1)}}{2}$
c
(c)$f(x) = \frac{x}{{1 + {x^2}}}$, $\therefore \,\,\,I = \int_{}^{} {f(x)} = \int_{}^{} {\frac{x}{{1 + {x^2}}}\,dx} $
Put $1 + {x^2} = t$

$\Rightarrow 2x\,dx = dt \Rightarrow x\,dx = dt2$
$\therefore \,\,\,I = \frac{1}{2}\int_{}^{} {\frac{{dt}}{t} = \frac{1}{2}\log t + c} $; $I = \frac{1}{2}\log (1 + {x^2}) + c$.

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

Find the value of the following determinant: $\begin{vmatrix}\displaystyle \frac{-4}{7} &\text{amp; } \displaystyle \frac{-6}{35}\\ 5 &\text{amp; } \displaystyle \frac{-2}{5}\end{vmatrix}$
  1. $\displaystyle \frac{15}{34}$
  2. $\displaystyle \frac{32}{45}$
  3. $\displaystyle \frac{25}{33}$​
  4. $\displaystyle \frac{38}{35}$​
→
If the system of linear equations $x + 2ay + az = 0$ $x + 3by + bz = 0$ $x + 4cy + cz = 0$ has a non-zero solution, then $a, b, c$
→
Let $\vec{a}$ and $\vec{b}$ be two vectors. Let $|\vec{a}|=1,|\vec{b}|=4$ and $\vec{a} \cdot \vec{b}=2$. If $\vec{c}=(2 \vec{a} \times \vec{b})-3 \vec{b}$, then the value of $\overrightarrow{ b } \cdot \overrightarrow{ c }$ is
→
Of the three independent events $E_1, E_2$ and $E_3$, the probability that only $E_1$ occurs is $\alpha$,only $E_2$ occurs is $\beta$ and only $E_3$ occurs is $\gamma$. Let the probability $p$ that none of events $E_1, E_2$ or $E_3$ occurs satisfy the equations ( $\alpha$ $-2 \beta) p=\alpha \beta$ and $(\beta-3 \gamma) p=2 \beta \gamma$. All the given probabilities are assumed to lie in the interval $(0,1)$.

Then $\frac{\text { Probability of occurrence of } E_1}{\text { Probability of occurrence of } E_3}=$

→
The corner points of the feasible region determined by the system of linear constraints are (0, 10), (5, 5), (15, 15), (0, 20). Let Z = px + qy, where p, q > 0. Condition on p and q so that the maximum of Z occurs at both the points (15, 15) and (0, 20) is:
→
If $\sin\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\cos\text{x}=\text{x}\sin\text{x},$ then $(\text{y}-1)\sin\text{x}=$
  1. $\text{c}-\text{x}\sin\text{x}$
  2. $\text{c}+\text{x}\cos\text{x}$
  3. $\text{c}-\text{x}\cos\text{x}$
  4. $\text{c}+\text{x}\sin\text{x}$
→
The function $f(x) = 2{x^3} - 3{x^2} - 12x + 4$ has
→
The value of $\int\limits_0^2 {\frac{{dx}}{{{{(1 - x)}^2}}}} $ is
→
$\int_0^1 {\frac{{{e^x}(x - 1)}}{{{{(x + 1)}^3}}}\,dx = } $
→
If $f(x) = \sin x - \cos x,$ the function decreasing in $0 \le x \le 2\pi $ is
→