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# Stuck on integration + rep watch

1. Hi,

Please can someone explain how the integral of:

(4-2u) / (4+u^2) = 2tan^-1(u/2) - ln(4+u^2)

I've tried by parts, looking for the log rule, but cant spot what i'm doing wrong.

Thanks alot.
2. (Original post by Crucio404)
Hi,

Please can someone explain how the integral of:

(4-2u) / (4+u^2) = 2tan^-1(u/2) - ln(4+u^2)

I've tried by parts, looking for the log rule, but cant spot what i'm doing wrong.

Thanks alot.
This shows how http://www.wolframalpha.com/input/?i...1%2Bx%29%5E0.5
3. Damn, partial fractions. Bad me.
4. Isn't it a substitution I like u/4=tan(theta).
5. (4-2u) / (4+u^2)
4/ (4+u^2) + (-2u)/ (4+u^2)

For: (-2u)/ (4+u^2) * du use the substitution v =4+u^2 => dv/2u=du
=> (-2u)*(dv/2u)/ (v) = -1/v * dv = -lnv +C= -ln(4+u^2) +C

The first integral is in the formula book. Hope that is clear.

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