Can't do this integral

How would I go about integrating $\dfrac {1}{\sqrt{x^2+1}}$ ?

Something to stay away from at A level. Like people have said, you need to know about hyperbolic functions to stand a chance of integrating this. Is it possible you have read the question wrong and you want to integrate 1/sqrt(1-x^2)

$\int \dfrac {1}{\sqrt{1+x^2}} dx [br][br][br][br]Let x = coshu; dx = sinhu du[br][br][br][br]= \int \dfrac {sinhu du}{\sqrt{1+cosh^2u}}[br][br]= u + C[br][br]arccosh x + C[br][br]$

Is this correct?

I realise I might've come off as gruff, but I'd like to stress I'm not What's contour integration? I wikipedia'd it but don't quite any of the terminology.

$\int \dfrac {1}{\sqrt{1+x^2}} dx [br][br][br][br]Let x = coshu; dx = sinhu du[br][br][br][br]= \int \dfrac {sinhu du}{\sqrt{1+cosh^2u}}[br][br]= u + C[br][br]arccosh x + C[br][br]$

Is this correct?

I realise I might've come off as gruff, but I'd like to stress I'm not What's contour integration? I wikipedia'd it but don't quite any of the terminology.

$\dfrac {1}{\sqrt{x^2+1}} = \dfrac{\dfrac{1}{\sqrt{x^2+1}} (x+\sqrt{x^2+1})}{x+\sqrt{x^2+1}}$
$= \dfrac{\dfrac{x}{\sqrt{x^2+1}} + 1}{x+\sqrt{x^2+1}}$
Which is a simple log integral

$\dfrac {1}{\sqrt{x^2+1}} = \dfrac{\dfrac{1}{\sqrt{x^2+1}} (x+\sqrt{x^2+1})}{x+\sqrt{x^2+1}}$
$= \dfrac{\dfrac{x}{\sqrt{x^2+1}} + 1}{x+\sqrt{x^2+1}}$
Which is a simple log integral

Not sure if the alcohol is affecting my maths ability now or not anymore, but I tried TS's problem with a trig substitution and it didn't work . Go for the hyperbolic.

You are pretty close, but cosh^2 u + 1 does not equal sinh^2 u. The Pythagorean identity is different in terms of hyperbolic functions (try another substitution). Do you know what complex numbers are? I think youll stuggle to understand contour integration without understanding of complex numbers.

oops

$\int \dfrac {1}{\sqrt{x^2 + 1}} dx[br][br]Let x = sinhu; dx = coshu du[br][br]\int \dfrac {coshu}{\sqrt{sinh^2u + 1}} du[br][br]arsinhx + C[br][br]$

Just basic knowledge, that $i = \sqrt{-1}$ and that you can treat it as a vector (the reals and complexes form a plane) and expressing it as r(cosx + isinx)

Congrats, your answer is right. Also use the log method as suggested by another poster to prove an interesting relationship between logs and hyperbolic functions.

As for the contour integrals, ill leave it to someone else to explain, because its exactly the sort of thing I'm likely to botch up.

Oooooooookaaaaaaay how about

$\int \sqrt{sinx} dx$


What am I doing not doing FM? I don't know...

Jesus, thats a horrible one.

As far as I know you can't integrate that in terms of normal functions.

Oooooooookaaaaaaay how about

$\int \sqrt{sinx} dx$


What am I doing not doing FM? I don't know...

That one has something to do with elliptic integrals.

What's that?

...

That one has something to do with elliptic integrals.

Wari wari, I don't have any experience with elliptic integrals. I entered the integral into Wolfram and that's what it came up with.