I just found out I forgot how integration works. Turns out that tomorrow I have an exam about it...

Making OP's integral easier:
[eqn]\int {{x+2}\over{\sqrt{x+1}}}dx=\int {(x-1)+3\over{\sqrt{x+1}}}=\int (x-1)^{0.5}dx+\int6\cdot{(x+1)^{-0.5}}dx[/eqn]

shoudl read 3, not 6

[eqn]\int {{x+2}\over{\sqrt{x+1}}}dx=\int {(x-1)+3\over{\sqrt{x+1}}}=\\
=\int (x-1)^{0.5}dx+3\int{(x+1)^{-0.5}}dx=\\
=\int u^{0.5}du+3\int v^{-0.5}dv=\\
=[{{1}\over{1.5}}u^{1.5}+{6}v^{0.5}+c]=\\
=[{{1}\over{1.5}}(x-1)^{1.5}+{6}(x+1)^{0.5}+c][/eqn]

hopefully got it right

Are Wolfram Alpha Pro's practice problems worth shit?

I thought only brainlets did manual labor.

To do [math]\int \frac{x+2}{\sqrt{x+1}}dx[/math]
Us the substitution [math]u=\sqrt{x+1}[/math]

This means that [math]\frac{d}{dx}u = \frac{du}{dx} = \frac{d}{dx} \frac{1}{\sqrt{x+1}}[/math]

Solving that derivative gives [math]\frac{d}{dx} (x+1)^\frac{1}{2} = \frac{1}{2}(x+1)^{-\frac{1}{2}}=\frac{1}{2\sqrt{x+1}}=\frac{1}{2u}[/math]

This means [math]\frac{du}{dx} = \frac{1}{2u} \implies dx=2u\ du[/math]

Substituting in [math]u=\sqrt{x+1}[/math] gives [math]\int \frac{u^2+1}{u} dx = \int (u+\frac{1}{u})(2u\ du) = 2\int (u^2+1)du = 2(\int u^2\ du + \int 1\ du)[/math]

That's a pretty simple integral, giving [math]2(\frac{u^3}{3}+u)+C[/math]

Undoing our subtitution we get [math]2\frac{\sqrt{x+1}^3}{3} + \sqrt{x+1} + C[/math]

Simplifing this gives the final result that
[math]\int \frac{x+2}{\sqrt{x+1}}dx= \sqrt{x+1}(\frac{2(x+1)}{3} +1) + C[/math]

I already went through that though, now I'm stuck with something that probably is retarded but I can't manage to get through my head:

[math]\int_{1}^{+\infty} \frac{1}{x(x^2+3)}dx[/math]

Confuses the hell out of me because I don't know how to do that partial fraction, I've tried a few things already. Hopefully I'll manage in a few hours, when I have the exam

HOLY SHIT THE RUSH. I GOT IT GUYS, I GOT IT, EUREKA GODDAMIT (TURNS OUT I WAS DOING PARTIAL FRACTIONS WRONG THE WHOLE TIME)
FUCK YEAH FULL CAPS
I wish you guys could feel my excitement, thank you all

just like babby when he learned of GT racers
brings a tear to my eye

it's just the opposite of derivation. think about how to find a tangent line on a curve and do the opposite.

u=x+1, du=dx. x=u-1. int[ (u-1)/u ]du + 2*int[ 1/u ]du = int[ u/u - 1/u]du + 2* int[ 1/u ]du = (x+1) - ln(x+1) + 2*ln(x+1).

oh woops, i forgot sqrt. so those u are actually u^(-1/2) and int[ u^(-1/2) ]du = 2*u^(1/2) = 2*sqrt(u).

Does anyone know formally why fiddling with the dx in substitution works?

[math]\int_{a}^{b} f(\varphi(t)) \cdot \varphi'(t)\,\mathrm{d}t = [/math]
[math]\int_{\varphi(a)}^{\varphi(b)} f(x)\cdot \frac_{\varphi(t)\cdot \mathrm{d}x}^{\mathrm{d}t}\,\mathrm{d}t = [/math]
[math]\int_{\varphi(a)}^{\varphi(b)} f(x)\,\mathrm{d}x[/math]

This seems hand-wavy.

Does anyone know formally why fiddling with the dx in substitution works?
[math]\int_{a}^{b} f(\varphi(t)) \cdot \varphi'(t)\,\mathrm{d}t =[/math]
[math]\int_{\varphi(a)}^{\varphi(b)}f(x)\cdot\frac{\varphi(t)\cdot\mathrm{d}x}^{\mathrm{d}t}\,\mathrm{d}t=[/math]
[math]\int_{\varphi(a)}^{\varphi(b)} f(x)\,\mathrm{d}x[/math]

>This seems hand-wavy.

\int_{\varphi(a)}^{\varphi(b)}f(x)\cdot\frac{\varphi(t)\cdot\mathrm{d}x}{\mathrm{d}t}\,\mathrm{d}t=

[math]\int_{\varphi(a)}^{\varphi(b)}f(x)\cdot\frac{\varphi(t)\cdot\mathrm{d}x}{\mathrm{d}t}\,\mathrm{d}t=[/math]