MCQ
$\int {\frac{{\cos x + x\sin x}}{{x(x - \cos x)}}dx = } $
- A$\log |x(x - \cos x)| + c$
- ✓$\log \left| {1 - \frac{{\cos x}}{x}} \right| + c$
- C$\log \left| {\frac{x}{{x - \cos x}}} \right| + c$
- DNone of these
Put $1-\frac{\cos x}{x}=t$
$-\left[\frac{-x \sin x-\cos x}{x^{2}}\right] d x=d t$
$\frac{x \sin x+\cos x}{x^{2}} d x=d t$
$\int \frac{\mathrm{dt}}{\mathrm{t}}$
$\ln t+c$
$ = \ln \left| {1 - \frac{{\cos {\rm{x}}}}{{\rm{x}}}} \right| + {\rm{c}}$
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$[3,5]$
$[-1,1]$
$\Big[-\sqrt5,-\sqrt3\Big]\cup\Big[\sqrt3,\sqrt5\Big]$
$\Big[-\sqrt5,-\sqrt3\Big]\cap\Big[\sqrt3,\sqrt5\Big]$
The general solution of
$\frac{\text{dy}}{\text{dx}}=2\text{x}\ \text{e}^{\text{x}^{2}-\text{y}}$ is:$\text{e}^{\text{x}^{2}-\text{y}}=\text{C}$
$\text{e}^{-\text{y}}+\text{e}^{\text{x}^{2}}=\text{C}$
$\text{e}^{\text{y}}=\text{e}^{\text{x}^{2}}+\text{C}$
$\text{e}^{\text{x}^{2}+\text{y}}=\text{C}$