I've looked it up on the internet but I'm having trouble as to how to proceed using Euler's substitution.
For example, how does one solve the following integrals using Euler's substitution:
- $\int_1^2\dfrac{dx}{x+\sqrt{x^2+1}}$
and
- $\int_0^{\infty} \dfrac{xdx}{\sqrt{x^5+1}}$
For the first one I tried the following:
$\sqrt{x^2+1} = x+t \implies x= \dfrac{1-t}{2}$
Therefore, $x+\sqrt{x^2+1} = 2-t^2$
We have $dx=-tdt$
Therefore: $\int_1^2\dfrac{dx}{x+\sqrt{x^2+1}} = -\int_1^2 \dfrac{tdt}{2-t^2} = \dfrac{1}{2}[log(2-t^2)]^2_1 $ Which is obviously wrong becaus $log(-2)$ doesn't exist.
For the first problem:
$$\int_1^2\dfrac{dx}{x+\sqrt{x^2+1}}$$
$$x+\sqrt{x^2+1} = t \implies \sqrt{x^2+1} = -x+t \implies x= \dfrac{t^2-1}{2t}$$
Calculate the derivative $dx$:
$$dx=\dfrac{t^2+1}{2t^2}dt$$
Substitution of $dx$ and $\int_1^2\frac{dx}{x+\sqrt{x^2+1}}$:
$$\int_1^2\dfrac{dx}{x+\sqrt{x^2+1}} = \int_1^2 \dfrac{t^2+1}{2t^2}*\dfrac{1}{t}dt = [\dfrac{1}{2}log(t)-\dfrac{1}{4}t^{-2}]_1^2$$
This is equal to:
$$\left[\dfrac{1}{2}log(x+\sqrt{x^2+1} )-\dfrac{1}{4}\times\dfrac{1}{(x+\sqrt{x^2+1} )^{2}}\right]_1^2$$