Anti-diff(sqrt(x^2+1)/x)dx from 1 to 2

[cos(e)]

Anti-diff(sqrt(x^2+1)/x)dx from 1 to 2

I have tried so many formulas and spent so much time on this and can't find a solution with working out, can someone please help me out.

 

[HallsofIvy]

Anti-diff(sqrt(x^2+1)/x)dx from 1 to 2

I have tried so many formulas and spent so much time on this and can't find a solution with working out, can someone please help me out.

Problems with a square inside a square root always make me think immediately of a trig substitution. $sin^2(\theta)+ cos^2(\theta)= 1$ and, dividing by $cos(\theta)$, $tan^2(\theta+ 1= sec^2(\theta)$. That tells us that if we let $x= tan(\theta)$, $\sqrt{x^2+ 1}= \sqrt{tan^2(\theta)+ 1}= sec(\theta)$. Of course, $dx= sec^2(\theta)d\theta$ so the integral becomes
[tex]\int \frac{sec(\theta)}{tan(\theta)}sec^2(\theta)d\theta[/itex]<br /> <br /> Now convert everything to sine and cosine: $sec(\theta)= 1/cos(\theta)$ and $tan(\theta)= sin(\theta)/cos(\theta)$[/tex]

 

[cos(e)]

Ok thanks for the fast reply, and i got down to anti-diff(1/(sinu*(cosu)^2)du), which i can't integrate any better than the first question i posted, can u point me in the right direction?

 

[snipez90]

Well you could write 1 as $$sin^2(\theta)+cos^2(\theta)$$ and then it will come down to finding the integral of $$sec(\theta)tan(\theta)$$ plus the integral of $$csc(\theta)$$.

Now the first is well-known and has antiderivative $$sec(\theta)$$ so that's not a problem. It's not too hard to derive the antiderivative for $$csc(\theta)$$ but you could always look it up.

 

[Defennder]

Use integration by parts. Rewrite it as sec^2 u csc u before doing so.

 

[HallsofIvy]

Ok thanks for the fast reply, and i got down to anti-diff(1/(sinu*(cosu)^2)du), which i can't integrate any better than the first question i posted, can u point me in the right direction?

Since that involves an odd power of sine, there is a standard method for it. Here the sine is in the denominator but we can multiply both numerator and denominator by sin(x) to get
[tex]\int \frac{sin u}{sin^2 u cos^2 u}du= \int \frac{sin u}{(1- cos^2u)cos^2 u}du[/itex]<br /> Now let y= cos u so dy= -sin u du or sin u du= -y dy. The integral becomes<br /> $$\int \frac{dy}{y(1-y^2)}= \int \frac{dy}{y(1-y)(1+y)}$$<br /> which can be done by "partial fractions".[/tex]