Integration methods: euler's substitution
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.
$\endgroup$1 Answer
$\begingroup$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:
$$\left[\dfrac{1}{2}log(x+\sqrt{x^2+1} )-\dfrac{1}{4}\times\dfrac{1}{(x+\sqrt{x^2+1} )^{2}}\right]_1^2$$
$\endgroup$ 5
