<s>[INST] Solve definite integral $\int_{\frac{\sqrt2}2}^{\sqrt{3}} \dfrac{dx}{x\sqrt{x^2+1}}$ by variable substitution I have this integral
$$\int_{\sqrt{2}/2}^{\sqrt{3}} \dfrac{dx}{x\sqrt{x^2+1}}$$
I have to solve it by substituting x.
I believe that the easiest way would be to substitute with $t = \sqrt{x^2+1}$. The other ways I tried didn't help me much.
so.. let $t = \sqrt{x^2+1}$. Suppose $\sqrt{6}/2 \leq t \leq 2$. In this case, we can say that $x = \varphi(t) = \sqrt{t^2-1}$. $\varphi(\sqrt{6}/2) = \sqrt{2}/2$ and $\varphi(2) = \sqrt{3}$.
Now we can substitute.
$$\int_{\sqrt{6}/2}^{2} {\dfrac{1}{\varphi(t)\sqrt{\varphi^2(t) + 1}}\varphi'(t)dt} = 
\int_{\sqrt{6}/2}^{2} {\dfrac{dt}{t\sqrt{t^2-1}}}$$
But I'm stuck here. Actually the indefinite integral of the integrand is equal to $-\ln{\sqrt{t^2-1}}$, but then by using the fundamental theorem of calculus I get a wrong result. What am I doing wrong?
The right answer is $\ln{\dfrac{3+\sqrt{6}}{3}}$. Thanks in advance.
 [/INST] Note that, with the substitution $t=\sqrt{x^2+1}$
$$ x = \sqrt{t^2-1}, \>\>\>\>\>dx = \frac t{\sqrt{t^2-1}}dt$$
Then, the integral becomes
$$\int_{\frac{\sqrt{2}}2}^{\sqrt{3}} \dfrac{1}{x\sqrt{x^2+1}}dx
= \int_{\sqrt{\frac32}}^{2} \dfrac{1}{t^2-1}dt
= \frac12 \ln\frac{t-1}{t+1}\bigg|_{\sqrt{\frac32}}^{2}\\
= \frac12\ln\left( \frac13\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}\right)=\ln\frac{3+\sqrt6}3
$$
</s>