$$ \int \sqrt{x^2+x+1} d x $$
$\frac{(2 x+1)}{4} \sqrt{x^2+x+1}+\frac{3}{8} \sinh ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+C$
$\frac{x+1}{4} \sqrt{x^2+x+1}+\frac{3}{8} \sinh ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+C$
$\frac{x+1}{4} \sqrt{x^2+x+1}-\frac{3}{8} \sinh ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+C$
$\frac{(2 x+1)}{4} \sqrt{x^2+x+1}-\frac{3}{8} \sinh ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+C$
If $k \in N$, then $\lim\limits_{n \rightarrow \infty}\left[\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\ldots .+\frac{1}{k n}\right]=$
$\log (k+1)$
$\log k$
$\log (k+5)$
$\log (k+1)-\log 6$
$$ \int_{-1}^4 \sqrt{\frac{4-x}{x+1}} d x= $$
0
$\frac{\pi}{2}$
$\frac{3 \pi}{2}$
$\frac{5 \pi}{2}$
$$ \int_0^{\pi / 4} \frac{\cos ^2 x}{\cos ^2 x+4 \sin ^2 x} d x= $$
$\frac{\pi}{2}-\frac{1}{3} \tan ^{-1} 2$
$-\frac{\pi}{4}-\frac{4}{3} \tan ^{-1} 2$
$\frac{\pi}{6}+\frac{2}{3} \tan ^{-1} 2$
$-\frac{\pi}{12}+\frac{2}{3} \tan ^{-1} 2$
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