If $P(\alpha, \beta)$ is a point on the curve $9 x^2+4 y^2=144$ in the first quadrant and the minimum area of the triangle formed by the tangent of the curve at $P$ with the coordinate axis is $S$, then
$S=\sqrt{\alpha \beta}$
$S=\alpha \beta$
$S=2 \sqrt{\alpha \beta}$
$S=2 \alpha \beta$
$$ \int(\log 2 x)^3 d x= $$
$x\left((\log 2 x)^3-3(\log 2 x)^2+6(\log 2 x)-6\right]+C$
$\frac{x}{4}\left[4(\log 2 x)^3-6(\log 2 x)^2+6(\log 2 x)-3\right]+C$
$\frac{x}{2}\left[(\log 2 x)^3-3(\log 2 x)^2+3(\log 2 x)-6\right]+C$
$x\left[(\log 2 x)^3-6(\log 2 x)^2+18(\log 2 x)-54\right]+C$
$$ \int \frac{x+1}{(x-2) \sqrt{1-x}} d x= $$
$\log (x+1)-\log (x-2) \sqrt{1-x}+C$
$\log (x-2) \sqrt{1-x}+C$
$6 \tan ^{-1} \sqrt{1-x}-2 \sqrt{1-x}+C$
$4 \tan ^{-1} \sqrt{1-x}-2 \sqrt{1-x}+C$
$$ \int \frac{1}{1+x+x^2} d x= $$
$\frac{2}{\sqrt{3}} \log \left(\frac{2 x+1+\sqrt{3}}{2 x-1-\sqrt{3}}\right)+C$
$\frac{1}{\sqrt{3}} \log \left(\frac{2 x+1-\sqrt{3}}{2 x+1+\sqrt{3}}\right)+C$
$\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+C$
$\frac{2}{\sqrt{5}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{5}}\right)+C$
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