<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] Let $$ \mathscr{I} = \int \dfrac{dx}{x\sqrt{x^2+1}}$$
Put $ x = \dfrac1t \implies dx = -\dfrac{dt}{t^2}$
$$ \begin{align} \mathscr{I} &\ =  -\int t\dfrac{1}{\sqrt{\frac{1}{t^2}+1}} \dfrac{dt}{t^2} \\ &\ =- \int\dfrac{dt}{\sqrt{t^2+1}} \\ &\ = - \ln \left| t + \sqrt{t^2+1} \right| + C
     \end{align} $$
Now, substitute the changed limits
$$ \begin{align} \int_{\frac{\sqrt{2}}{2}}^{\sqrt{3}} \dfrac{dx}{x\sqrt{x^2+1}} &\ =   \ln \left| t + \sqrt{t^2+1} \right|_{\frac{1}{\sqrt3}} ^\sqrt2 \\  &\ = \ln \dfrac{\sqrt2 + \sqrt3}{\frac{1}{\sqrt3} + \frac{2}{\sqrt3} } \\ &\ = \ln{\dfrac{3+\sqrt{6}}{3}} \end{align} $$
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