MCQ
The integral $16 \int \limits_1^2 \frac{d x}{x^3\left(x^2+2\right)^2}$ is equal to
- A$\frac{11}{6}+\log _e 4$
- B$\frac{11}{12}+\log _e 4$
- C$\frac{11}{12}-\log _{ e } 4$
- ✓$\frac{11}{6}-\log _e 4$
$=16 \int \limits_1^2 \frac{d x}{x^3 x^4\left(1+\frac{2}{x^2}\right)^2}$
Let, $1+\frac{2}{x^2}=t \Rightarrow \frac{-4}{x^3} d x=d t$
$I=-4 \int \limits_3^{\frac{3}{2}} \frac{d t}{\left(\frac{2}{t-1}\right)^2 t^2}$
$I=-4 \int \limits_3^{\frac{3}{2}}\left(\frac{t-1}{2}\right)^2 \frac{d t}{t^2}$
$I=-\frac{4}{4} \int \limits_3^{\frac{3}{2}}\left(1-\frac{2}{t}+\frac{1}{t^2}\right) d t$
$I=-1\left[t-2 \ell n|t|-\frac{1}{t}\right]_3^{\frac{3}{2}}$
$I=-1\left[\left(\frac{3}{2}-2 \ell n \frac{3}{2}-\frac{2}{3}\right)-\left(3-2 \ell n 3-\frac{1}{3}\right)\right]$
$I=-1\left[2 \ell n 2-\frac{11}{6}\right]$
$I=\frac{11}{6}-\ell n 4$
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If $I_1 = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}} \, f (\tan\, \theta + \cot\, \theta )\cdot sec^2\, \theta\, d\, \theta$ &
$I_2 = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}} \, f (\tan\, \theta + \cot\, \theta )\cdot cosec^2\, \theta\, d \, \theta$ ,
then the ratio $\frac{{{I_1}}}{{{I_2}}}$ :
and $g(x)=\left(x-\frac{1}{2}\right)^{2}, x \in R .$ Then the area (in sq. units) of the region bounded by the curves, $y=f(x)$ and $y=g(x)$ between the lines, $2 \mathrm{x}=1$ and $2 \mathrm{x}=\sqrt{3},$ is