If the lengths of the tangent, subtangent, normal and subnormal for the curve $y=x^2+x-1$ at the point $(1,1)$ are $a, b, c$ and $d$ respectively, then their increasing order is
$b, d, a, c$
b, a, c, d
$a, b, c, d$
$b, a, d, c$
$$ \int \frac{x+1}{x^3-1} d x= $$
$\frac{1}{3} \log \left(\frac{x+1}{x^2+x+1}\right)+C$
$\frac{1}{3} \log \left(\frac{(x-1)^2}{x^2+x+1}\right)+C$
$\frac{1}{3} \log \left(\frac{x-1}{x^2+x+1}\right)+C$
$\frac{1}{3} \log \left(\frac{(x+1)^2}{x^2-x+1}\right)+C$
$$ \int \frac{x^4-16 x^2+2 x+8}{x^3-4 x^2+2} d x= $$
$\frac{x^2+8 x+C}{2}$
$x^2+8 x+C$
$x^3-4 x+C$
$\frac{x^2-8 x+C}{2}$
$$ \int \frac{\sec ^2 x}{(\sec x+\tan x)^{\frac{5}{2}}} d x= $$
$-\frac{(\sec x+\tan x)^{\frac{5}{2}}}{5}-\frac{(\sec x+\tan x)^{\frac{7}{2}}}{7}+C$
$-\frac{(\sec x-\tan x)^{\frac{5}{2}}}{5}-\frac{(\sec x-\tan x)^{\frac{7}{2}}}{7}+C$
$-\frac{(\sec x+\tan x)^{\frac{3}{2}}}{3}-\frac{(\sec x+\tan x)^{\frac{7}{2}}}{7}+C$
$-\frac{(\sec x-\tan x)^{\frac{3}{2}}}{3}-\frac{(\sec x-\tan x)^{\frac{7}{2}}}{7}+C$
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