Question

Two integral problems: Explanations and steps please!

Integral of (s+1)/(4-s^2)^(1/2)
::I don't know how to deal with the + 1 part.
Integral of (tan^-1 x)/(1+x^2)
::I've never dealt with one of these. I tried integration by parts but is stuck with the inverse tangent.

Answer 1

http://www.numberempire.com/cgi-bin/render2.cgi?%5Cnocache%20%5CLARGE%20%5Cint%7B%7B%7Bx%2B1%7D%5Cover%7B2%5C%2C%5Cleft%284-x%5E2%5Cright%29%7D%7D%7D%5C%2Cdx%20%3D%20-%7B%7B%5Clog%20%5Cleft%28x%2B2%5Cright%29%2B3%5C%2C%5Clog%20%5Cleft%28x-2%5Cright%29%7D%5Cover%7B8%7D%7D

Answer 2

here are the solution

Answer 3

1st one you can write as \[\int\limits \frac{s}{\sqrt{4-s^2}}ds+\int\limits \frac{1}{\sqrt{4-s^2}}ds\]
Here the 1st one is easy and for the second one use substitution
\[s=2\sin(t)\]

Answer 4

2nd is \[\int\limits (\tan^{-1}x)(1+x)dx\] Use substitution \[t = \tan^{-1}x\]

Answer 5

sorry i misspelled is tan^(-1)x *(1+x^2)

Answer 6

second is a set up for u -sub, have
\[\int \frac{\tan^{-1}( x)}{1+x^2}dx\] but
\[u=\tan(x),du=\frac{dx}{1+x^2}\] and you get
\[\int u du\] ready to go

Answer 7

typo above, should have put
\[u=\tan^{-1}(x)\]

Answer 8

ah , I even wrote the problem wrong :D , Satellite is correct. But the main idea is the substitution
\[t = \tan^{-1}x\]

Answer 9

Oh, thanks. I'm getting stuck in so many exercises.

Answer 10

Wait, actually I still don't get the second one.
If \[u = \tan^-1 (x)\]
Won't it become \[\tan^-1 (\tan^-1 x)\]

And I don't get how that works because if u is that then \[du = 1/(x^2+1)\]
You put it in and it'll become \[\int\limits_{?}^{?}(1+x^2)/(1+x^2)dx\]
How does that work? How do pull it out?
It's \[\int\limits_{?}^{?} du/(x^2+1)\]

Answer 11

\[\ u = tan^{-1}x \]
\[\ du = \frac{1}{1+x^{2}}dx\]
You can now substitute the expression(1/(1+x^2))dx in the integral with du
If you don't see it right away, try doing this"
\[\ dx = (1+x^2)du\]
and now plug this into the integral and you will get"
\[\int \frac{u}{1+x^2}(1+x^2)du \]

Answer 12

Oh, I see. Thank you.