#Integration
85 messages · Page 1 of 1 (latest)
use intergral of 1/x=lnx
It’s such a trick u sub
$\sqrt{x+5}=u$
axkyn
$u^2=x+5$
axkyn
$U^2-5=x$
axkyn
axkyn
$\int 2u(2+u)^{-1}du$
axkyn
2 outside becos constant
then write u as u+2-2?
does that work?
Yeah that’s what I was thinking idk
Oh shoot
My bad, let me fix that
This is new to me hold on
Oh but wait it’s a product
The integral, would you use integration by parts?
After the u-sub?
Let me try that
$2\int \frac{u}{u+2}du$
axkyn
$-4\int \frac{1}{u+2}du$
axkyn
What did u do here?
$-4(ln|u+2|)+C$
axkyn
$\sqrt{x+5}=u$
axkyn
$-4(ln|\sqrt{x+5}+2}|)+C$
axkyn
Compile Error! Click the
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This is probably wrong
Can u tell me how above, u went from a constant of 2 outside the integral to -4, and how the u for the numerator became a 1?
$2 \int \frac{u+2-2}{u+2}du$
axkyn
$2\int \frac{-2}{u+2}du$
axkyn
$-4 \int \frac{1}{u+2} du$
axkyn
@versed estuary
Ok that makes sense, thank you
Yeah, it’s my first year of AP calc.
What’s your age
Dang
Oh your older then me
Thanks now i got it!
I´m 19, first year in college. We´re having calc test on thursday.
Nice
You all are older than me 
You got it
(u+2-2)/(u+2) = 1 - 2/(u+2)
Oh shi
Thanks
$2(u-2\int \frac{1}{u+2} du$
axkyn
$2u-4 ln|u+2|+C$
axkyn
@glad burrow silly mistake by me
It's fine just be careful
I was walking on a trail 
yeah no thats wrong
Yeah ik