$$ \int \frac{x+1}{(x-2) \sqrt{1-x}} d x= $$
$\log (x+1)-\log (x-2) \sqrt{1-x}+C$
$\log (x-2) \sqrt{1-x}+C$
$6 \tan ^{-1} \sqrt{1-x}-2 \sqrt{1-x}+C$
$4 \tan ^{-1} \sqrt{1-x}-2 \sqrt{1-x}+C$
$$ \int \frac{1}{1+x+x^2} d x= $$
$\frac{2}{\sqrt{3}} \log \left(\frac{2 x+1+\sqrt{3}}{2 x-1-\sqrt{3}}\right)+C$
$\frac{1}{\sqrt{3}} \log \left(\frac{2 x+1-\sqrt{3}}{2 x+1+\sqrt{3}}\right)+C$
$\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+C$
$\frac{2}{\sqrt{5}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{5}}\right)+C$
If $\int \frac{d x}{(x \tan x+1)^2}=f(x)+C$, then $\lim\limits_{x \rightarrow \frac{\pi}{2}} f(x)=$
$\frac{\pi}{2}$
$\frac{2}{\pi}$
$\frac{1}{\pi}$
$\infty$
$$ \int \sin ^3 x \cos ^2 x d x= $$
$\frac{\sin ^4 x \cos x}{5}-\frac{\sin ^2 x \cos x}{15}-\frac{2 \cos x}{15}+C$
$-\frac{\sin ^4 x \cos x}{5}-\frac{\sin ^2 x \cos x}{15}+\frac{2 \cos x}{15}+C$
$\frac{\sin ^4 x \cos ^{\prime} x}{5}-\frac{\sin ^2 x \cos x}{15}+\frac{2 x}{15}+C$
$\frac{\sin ^4 x \cos x}{5}+\frac{\sin ^2 x \cos x}{3}-\frac{2 x}{15}+C$
AP EAPCET Papers
All year-wise previous year question papers