C-Could one of you help me with this?
It looks like we could use integration by parts here after a quick u substitution, by if I let
u = 1/x,
-du = (1/x^2)dx
But then we're left with the pesky x^-1 that won't simplify upon taking its derivative
C-Could one of you help me with this?
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You've gotta be joking.
?
Can you give a hint at least?
Wat
I am suck at chemistry
This is calc 2 m8
At quick glance you should first integrate by parts and after that substitute u=1/x.
This sounds logical to me, but I'm rusty.
wut is this
If you substitute ir will simplify to integral -ue^u du where u = 1/x
CRINGEEEE
Aftrr that its integration by parts to reduce -u to -1
(x−1)e^1/x/x+C
So you have -ue^u + e^u = e^u(1-u)
((x-1)e^1)/x + c *
Ahh I see, we further substitute u for the left over 1/x term when fist doing the substitution. Brilliant! Thank you!
C-could you fuck off with your cancer anime uguu writing s-style?
Funny thing I havent officialy started integrals on my uni and yet I solved it in my head on a mobile. Youre welcome
Wow I don't remember any of this.
Calc 2 almost gave me PTSD though. Jesus, that was brutal.
Good work studying that stuff, user!
So. Solution: e^(1/x)*(x - 1)/x
1. Substitute:
U=1/x. du = -1/x^2
Gives: integral[ -(e^u)*u du]
2. Integration by parts
Integral[ -e^u*u du] = -e^u*u +integral[e^u du]
= e^u(u -1)
Aaaaand u= 1/x and reduce the expresion. But that should do it. Bit rusty too so hope it helps
Missed the constant in the solution .. just noticed.. damn those arbitrary constants
Oh dear.
e^1/x/x . Pure cancer
Impressive
How? Are you just so smart that simply knowing derivatives and the fact integration is the inverse to them that these results come naturally/ trivially to you?