_ ^ dx ( x^2 + 1)( x^2 + 4) = - Vidyadip

MCQ

$\int_{}^{} {\frac{{dx}}{{({x^2} + 1)({x^2} + 4)}} = } $

A
$\frac{1}{3}{\tan ^{ - 1}}x - \frac{1}{3}{\tan ^{ - 1}}\frac{x}{2} + c$
B
$\frac{1}{3}{\tan ^{ - 1}}x + \frac{1}{3}{\tan ^{ - 1}}\frac{x}{2} + c$
✓
$\frac{1}{3}{\tan ^{ - 1}}x - \frac{1}{6}{\tan ^{ - 1}}\frac{x}{2} + c$
D
${\tan ^{ - 1}}x - 2{\tan ^{ - 1}}\frac{x}{2} + c$

✓

Answer

Correct option: C.

$\frac{1}{3}{\tan ^{ - 1}}x - \frac{1}{6}{\tan ^{ - 1}}\frac{x}{2} + c$

c
(c)$\int_{}^{} {\frac{{dx}}{{({x^2} + 1)({x^2} + 4)}}} = \frac{1}{3}\left[ {\int_{}^{} {\frac{{dx}}{{{x^2} + 1}} - \int_{}^{} {\frac{{dx}}{{{x^2} + 4}}} } } \right]$
$ = \frac{1}{3}\left[ {{{\tan }^{ - 1}}x - \frac{1}{2}{{\tan }^{ - 1}}\frac{x}{2}} \right] + c = \frac{1}{3}{\tan ^{ - 1}}x - \frac{1}{6}{\tan ^{ - 1}}\frac{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 "MCQ" from 7.1 inderfinite integral→7.1 inderfinite integral — all question types→ગણિત ધોરણ 12 સાયન્સ question papers→Gujarat Board ધોરણ 12 સાયન્સ question papers→Gujarat Board question paper generator→

Similar questions

જો $y = x\log \left( {{x \over {a + bx}}} \right)$, તો ${x^3}{{{d^2}y} \over {d{x^2}}} = $

જો સદિશો $a,\,b,\,c$ માટે $[a\,b\,c\,] = 4$ ,તો $[a \times b\,\,b \times c\,\,c \times a]$ =

ધારોકે $y=y(x)$ એ વિકલ સમીકરણ $(x \cos x) d y+(x y \sin x+y \cos x-1) d x=0,0 < x < \frac{\pi}{2}$ નો ઉકેલ છે.જો $\frac{\pi}{3} y\left(\frac{\pi}{3}\right)=\sqrt{3}$ હોય, તો $\left|\frac{\pi}{6} y^{\prime \prime}\left(\frac{\pi}{6}\right)+2 y^{\prime}\left(\frac{\pi}{6}\right)\right|=.............$.

$\lim _{n \rightarrow \infty}\left(\frac{1}{1+n}+\frac{1}{2+n}+\frac{1}{3+n}+\ldots+\frac{1}{2 n}\right)$ મેળવો.

જો $A= \{1, 2, 3, 4\}$ અને સંબંધ $R : A \to A$ ; $R = \{ (1, 1), (2, 3), (3, 4), ( 4, 2) \}$ આપેલ હોય તો આપેલ પૈકી સત્ય વિધાન મેળવો.

વિકલ સમીકરણ $\frac{d^2 y}{d x^2}+x^2 \log \left(\frac{d y}{d x}\right)=y$ નું પરિમાણ ____________ છે.

વિધેય $f(x)\, = \left\{ \begin{array}{l}x + 2\,\,\,\,,\,\,\,1 \le x \le 2\\4\,\,\,\,\,\,\,\,\,\,\,,\,\,\,x = 2\\3x - 2\,\,,\,\,\,x > 2\end{array} \right.$ એ . . . બિંદુએ સતત થાય.

$I = \int\limits_0^{\pi /2} {\frac{{{{\left( {\sin x + \cos x} \right)}^2}}}{{\sqrt {1 + \sin 2x} }}\,dx = ........} $

વિધેય ${f}{\text{(x)}}\,\, = \,\int\limits_{\text{0}}^{\text{x}} {\frac{{{\text{sin}}\,{\text{t}}}}{{\text{t}}}} $ માટે, જ્યાં $\,{\text{x}}\,\, > \,{\text{0}}\,\, = \,\,......$

એક રેખા $X $ અને $Z$ અક્ષ સાથે $\theta$ તથા $Y$ અક્ષ સાથે $\beta$ ખૂણો બનાવે છે તથા $sin^2 \beta = 3 sin^2 \theta $ તો $ cos^2 \theta = ……$