how to integrate sqrt(x)/(1+x)?

Question

anonymous · Answer

try $u=\sqrt{x}$

agent0smith · Answer

$$\Large \int\limits_{}^{}\frac{ \sqrt x }{ 1+x } dx$$
Looks like you'll have to make a trig substitution as well as a regular substitution...idk exactly how it'll work out since i haven't tried it yet.

Maybe start with u = sqrtx, so u^2 = x, and 
du= dx/(2sqrtx) = dx/2u 
so dx = 2u du
$$\Large \int\limits_{}^{}\frac{ u }{ 1+u^2 }*2u du$$
idk if this is going anywhere... but you can use int. by parts from here i think.

anonymous · Answer

then $u^2=x, 2udu=dx$ and integrate 
$$\int \frac{u^2}{1+u^2}du$$

anonymous · Answer

oops i dropped a 2

anonymous · Answer

$$\Large 2\int\frac{ u^2 }{ 1+u^2 } du$$

dan815 · Answer

doh why did i delete that

anonymous · Answer

$\frac{u^2}{1+u^2}=\frac{u^2+1-1}{u^2+1}=1-\frac{1}{1+u^2}$

anonymous · Answer

now it should be easier

anonymous · Answer

Satellite73 how did you get that?

anonymous · Answer

don't need parts, although i guess it could work
$$\int \frac{du}{1+u^2}$$ is well known

anonymous · Answer

how did i get what?

isaiah.feynman · Answer

@satellite73 Good algebra usage there.

anonymous · Answer

lol thanks
you can also divide i guess, but it is easier to do it that way i think

agent0smith · Answer

you can also get this by doing polynomial long division, btw @ryanmk54  $$\Large \frac{u^2}{1+u^2}=\frac{u^2+1-1}{u^2+1}=1-\frac{1}{1+u^2}$$

isaiah.feynman · Answer

Wow..the algebra simplifies everything easily.

anonymous · Answer

$$u^2 /(1+u^2)$$ I got u over$$u/(u^2 +1)$$

anonymous · Answer

don't forget that $2udu=dx$ which is why you have $u^2$ up top and not $u$

anonymous · Answer

Did you then simplify by creatively adding 0? $$u^2 + 1 -1$$

agent0smith · Answer

^yes he did!

anonymous · Answer

yes, but that is because i have seen this gimmick many times
without it, you have to divide 
since you have $u^2$ up top and $u^2$ in the bottom, you know it will be $1+\frac{a}{1+u^2}$ so it is easier just to force it instead of diving using long division

anonymous · Answer

*dividing

anonymous · Answer

Thx agent0smith and satellite73

anonymous · Answer

Integral of the second term would just be ln(1-u^2)du?

agent0smith · Answer

^no, integral of 1/(1+x^2) = arctan x    ...iirc

anonymous · Answer

Your right thx