Can somoene help me to solve pls
#integral problem
hearty narwhal
austere tundraBOT
knotty junco
$\frac{1}{x^2\sqrt{x^2+1}} = \frac{\sqrt{x^2 + 1}}{x^2} - \frac{1}{\sqrt{x^2 + 1}}$
forest cobaltBOT
Daddy_314
knotty junco
You can then integrate by parts the second term
modern crown
how does one notice that besides "by inspection"
knotty junco
You don't have to
If you did not notice that, no problem
One thing that works with sums of squares under roots is hyperbolic trigonometry substitution
$sinh(u) = x$ \
$du cosh(u) = dx$ \
$x^2+1 = sinh^2(u)+1 = cosh^2(u)$ \
$$\int \frac{cosh(u)}{cosh(u)sinh^2(u)}du = -coth(u)
= -coth(sinh^{-1}(x))$$
And this has a very simple expression
forest cobaltBOT
Daddy_314
knotty junco
how does one notice that besides "by inspection"
You can also start by "rationalizing the denominator" by multiplying by sqrt(x^2+1)/sqrt(x^2+1)
You get
sqrt(x^2+1)/(x^2(x^2+1))
Start by finding partial fraction decomposition of 1/(x^2(x^2+1)) using the very standard methods, then multiply everything by sqrt(x^2+1)
knotty junco
Life is hard unfortunately
modern crown
ye
hearty narwhal
**Daddy\_314**
Could you develop a bit more your way of thinking here please ?
knotty junco
Which part
The decomposition as a sum ?
Or the integration by parts ?
The tricky part is the integration by parts
knotty junco
$sinh(u) = x$ \\
$du cosh(u) = dx$ \\
$x^2+1 = sinh^2(u)+1 = cosh^2(u)$ \\
$$...
You can use this method
If you are familiar with changes of variables
And if you know the hyperbolic cosine and sine functions
hearty narwhal
The décomposition as a sum pls
knotty junco
hearty narwhal
Haha i know but i did it alone.
And i just want to know if my answer is correct
Because i get this
knotty junco
What is this insect on the right
hearty narwhal
There is a minus sign somewhere
Oh yea that's right i forgot to write it
hearty narwhal
What is this insect on the right
It's the letters i use to say that it can be plus every number
knotty junco
$sinh(u) = x$ \\
$du cosh(u) = dx$ \\
$x^2+1 = sinh^2(u)+1 = cosh^2(u)$ \\
$$...
And this method here
Gives the same result
Because
$-coth(sinh^{-1}(x)) = -\frac{\sqrt{x^2+1}}{x^2}$
forest cobaltBOT
Daddy_314
knotty junco
fierce kayak
easy way of doing this would be taking x^2 common inside the square root and then taking it outside the square root, so u get 1/x^3 * sqrt(1 + x^-2) now just take the x^3 in the numerator so u get x^-3 / sqrt(1+x^-2) now put 1+x^-2 = t and the numerator is already a ready-made derivative, no need to use IBP
slow cape
guys there is an easier way to do this lol
factor x^2 from the root
you'll get x^-3/ sqrt(1 +x^-2)
then use u = 1 +x^-2
slow cape
easy way of doing this would be taking x^2 common inside the square root and the...
oh someone already said this
this is awkward now 💀