$$ \int \frac{x^4-16 x^2+2 x+8}{x^3-4 x^2+2} d x= $$
$\frac{x^2+8 x+C}{2}$
$x^2+8 x+C$
$x^3-4 x+C$
$\frac{x^2-8 x+C}{2}$
$$ \int \frac{\sec ^2 x}{(\sec x+\tan x)^{\frac{5}{2}}} d x= $$
$-\frac{(\sec x+\tan x)^{\frac{5}{2}}}{5}-\frac{(\sec x+\tan x)^{\frac{7}{2}}}{7}+C$
$-\frac{(\sec x-\tan x)^{\frac{5}{2}}}{5}-\frac{(\sec x-\tan x)^{\frac{7}{2}}}{7}+C$
$-\frac{(\sec x+\tan x)^{\frac{3}{2}}}{3}-\frac{(\sec x+\tan x)^{\frac{7}{2}}}{7}+C$
$-\frac{(\sec x-\tan x)^{\frac{3}{2}}}{3}-\frac{(\sec x-\tan x)^{\frac{7}{2}}}{7}+C$
$$ \int \frac{1}{\cos x}\left[\frac{1}{\sin x}-\frac{1}{\sin x+3 \cos x}\right] d x= $$
$\frac{1}{3} \log \left|\frac{\sin x}{\sin x+3 \cos x}\right|+C$
$\log \left|\frac{\cos x}{\sin x+3 \cos x}\right|+c$
$\frac{1}{3} \log \left|\frac{\cos x}{\sin x+3 \cos x}\right|+C$
$\log \left|\frac{\sin x}{\sin x+3 \cos x}\right|+c$
$$ \int \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right) d x= $$
$2\left[x \tan ^{-1} x-\log \sqrt{1+x^2}\right]+C$
$2 x \tan ^{-1} x+\log \sqrt{1+x^2}+C$
$x \tan ^{-1} x+\log \sqrt{1-x^2}+C$
$2\left[\tan ^{-1} x-\log \sqrt{1+x^2}\right]+C$
AP EAPCET Papers
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