How can I evaluate this integral?
$ \int_2^3 \frac{1}{\sqrt{x^2-x-2}} dx$
I know that this integral is convergent but I don't know how to evaluate it.
How can I evaluate this integral?
$ \int_2^3 \frac{1}{\sqrt{x^2-x-2}} dx$
I know that this integral is convergent but I don't know how to evaluate it.
$I = \displaystyle \int_2^3 \frac{dx}{\sqrt{x^2 - x -2}} = \displaystyle \int_2^3 \frac{dx}{\sqrt{ \left(x - \frac12 \right)^2 - \left(\frac{3}{2} \right)^2}}$ Now let $\left(x - \frac12 \right) = \frac32 \sec(\theta)$. This gives us $dx = \frac32 \sec(\theta) \tan(\theta) d \theta$. Hence, we get $I = \int_{0}^{\theta_1} \sec(\theta) d \theta = \log \left( \sec(\theta_1) + \tan(\theta_1) \right)$ where $\cos(\theta_1) = \frac35$. Hence, we get $I = \log \left(\frac53 + \frac43 \right) = \log(3)$
Put $(x^2-x)= (x-1/2)^2-1/4$ to obtain $\int_2^3 \dfrac{1}{\sqrt{x^2-x-2}}dx=\int_2^3 \dfrac{1}{\sqrt {(x-1/2)^2-9/4}}dx$ and do the change of variable $u=x-1/2$ then the integral becomes $\int_{3/2}^{5/2} \dfrac{1}{\sqrt {u^2-9/4}} du=...$ the rest is an exercise.
Since this is very likely a homework problem,
HINT: $x^2-x-2 = (x+1)(x-2)$. Thus $\int_2^3 \frac{\mathrm{d} x}{\sqrt{(x+1)(x-2)}} \stackrel{y=x+2}{=} \int_0^1 \frac{\mathrm{d} y}{\sqrt{(y+3) y}} \stackrel{y=u^2}{=} \int_0^1 \frac{2 \mathrm{d} u}{\sqrt{3+u^2}}$.