Integral of (2-u)/(1+u^2) du

Question

Integral of (2-u)/(1+u^2)  du

anonymous · Answer

$$\int\limits (2-u)/(1+u^2)   du$$

raden · Answer

use trig-subst it will work...

raden · Answer

do u know integration using trig-subs ?

anonymous · Answer

Not all that well.  I'm looking around online for some trig identities to work and no luck so far.

raden · Answer

ok, first let u=tan(theta), so du=....
for (1+u^2) it can be equal 1+(tan(theta))^2 = ....

anonymous · Answer

I'm seeing that if u = tan(theta), then du = sec^2(theta) 
How does this help?

raden · Answer

yes, u are right...
derivative of tan(theta) is sec^2(theta).
for the second, for 1+u^2 = ... ?

raden · Answer

put u=tan(theta)

anonymous · Answer

ok...  1+(tan(theta))^2

raden · Answer

ok, from identity trigonometry :
1+(tan(theta))^2 = ... ?

anonymous · Answer

1 + tan(theta))^2 = sec^2(theta)

raden · Answer

u are correct again 
dont forget, for 2-u, just put u=tan(theta)
so, subtitute of them to that problem , it should be ?

anonymous · Answer

$$\int\limits 2-\tan(\theta) d \theta$$

raden · Answer

yeps... just integrate that !

anonymous · Answer

$$∫2−\tan(θ)dθ = 2\theta-\ln(\sec \theta) +c$$

raden · Answer

for a moment, that's correct
but the answer shoulde be in "x", right ?

anonymous · Answer

well,  u  really in this problem

anonymous · Answer

$$2(\arctan u) - \ln(\sec(\arctan u))+c$$

raden · Answer

ops,, sorry into "u" i mean ...

raden · Answer

from u=tan(theta), it means 
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