Question

integral sqrt(x+1)/x^2 dx from 1 to infinite to compute and prove that it's convergent. for the convergence i know that if lim from x to inf from the value of the integral is a real number the integral converges. the computation is the issue here.

Answer 1

So you need to figure out whether the following integral converges:
\[\int_1^\infty \frac{\sqrt{x+1}}{x^2}~dx~~?\]

Answer 2

yes

Answer 3

A u-sub should work nicely here.

Let \(u=\sqrt{x+1}\), so that you get \(du=\dfrac{1}{2\sqrt{x+1}}dx\), or equivalently, \(dx=2u~du\).

From the initial substitution, you also know that \(u^2=x+1\), which tells you \(x=u^2-1\).

Keep track of your limits:
upper: \(x\to\infty~~\Rightarrow~~u\to\infty\).
lower: \(x=1~~\Rightarrow~~u=\sqrt{1+1}=\sqrt2\).

So you have
\[\large \int_\sqrt2^\infty \frac{u}{\left(1-u^2\right)^2}\bigg(2u~du\bigg)\\
\large 2\int_\sqrt2^\infty \frac{u^2}{\left(1-u^2\right)^2}~du\\
\]

Answer 4

Partial fractions from here.

Answer 5

You can simplify this somewhat with another substitution. It'd make the PFD easier on the eyes, but not necessary.

Answer 6

as in replace the u^2 w/ say v and go w/ partial fractions from there?

Answer 7

Something like that, yeah. It'd involve a square root, though. I'd stick with this first sub for now though.

Answer 8

btw, when you did the substitution you gave the integrand \[\frac{ u }{ (1-u^2)^2 }\]. wasn't it supposed to be \[\frac{ u }{ (u^2 - 1)^2 }\]?

Answer 9

did i miss something there?

Answer 10

Maybe some color coding will help. Just a sec.

Answer 11

Sorry, had some connection troubles.

Answer 12

Got it now. Re-did the entire integral myself now and figured out where it all fits. Ty very much.

Answer 13

\[\int_1^\infty\frac{\sqrt{x+1}}{x^2}~dx\]
\(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles
A u-sub should work nicely here.

Let \(\color{red}{u=\sqrt{x+1}}\), so that you get \(\color{blue}{du=\dfrac{1}{2\sqrt{x+1}}dx}\), or equivalently, \(\color{blue}{dx=2u~du}\).

From the initial substitution, you also know that \(\color{purple}{u^2=x+1}\), which tells you \(\color{purple}{x=u^2-1}\).

Keep track of your limits:
upper: \(x\to\infty~~\Rightarrow~~u\to\infty\).
lower: \(x=1~~\Rightarrow~~u=\sqrt{1+1}=\sqrt2\).

So you have
\[\large \int_\sqrt2^\infty \frac{u}{\left(1-u^2\right)^2}\bigg(2u~du\bigg)\\
\large 2\int_\sqrt2^\infty \frac{u^2}{\left(1-u^2\right)^2}~du\\
\]
\(\color{blue}{\text{End of Quote}}\)

\[\int_1^\infty \frac{\color{red}{\sqrt{x+1}}}{\color{purple}{x}^2}~\color{blue}{dx} =2\int_\sqrt2^\infty \frac{\color{red}{u}}{\left(\color{purple}{u^2-1}\right)^2}~\color{blue}{2u~du}=2\int_\sqrt2^\infty \frac{u^2}{\left(u^2-1\right)^2}~du\]

Answer 14

You're welcome