[\int\limits_{}^{}\frac{ e^{4/x} }{ x^{2} }] please help :)
\[\int\limits_{}^{}\frac{ e^{4/x} }{ x^{2} }\]
let u = 4/x , what is dx?
4lnx?
no, derivative of: u = 4/x
i know 1/x is lnx
its not 4 ln x?
he isn't asking you to integrate 1/x
not integration i am considering a u substitution, so we have to replace all the x parts appropriately if we let u = 4/x, and take the derivative of u with respect to x, what do we get?
i understand, it's 4/x wouldn't it be 4lnx
oh wow i was integrating
\[\frac{e^{4/x}}{x^2}dx\] \[\frac{e^{u}}{x^2}dx\] we need to replace the other x parts with u parts letting u = 4/x
latex seems to be on the fritx
**fritz
so it's deriv of 4x^-1 = -4x^-2
almost du = -4x^(-2) dx -du/4 = x^(-2) dx
ah i see now
e^u ---- du is our substitution -4
now, this is a constant and a rather simple e^u run up
so (-1/4)(e^4/x)+c?
take the derivative to chk
http://www.wolframalpha.com/input/?i=derivative++-e%5E%284%2Fx%29%2F4 it works :)
deriv of (-1/4)(e^4/x)+C = (-1/4)(e^4/x)(4/x)'=(-1/4)(e^4/x)(-4/x^2)=(e^4/x)/x^2 woooo
lol wolfram makes me sad cause the step by step explanations helped me out a lot during cal I, but i'm too poor to pay for it right now lol :(
my next questions are trig integrations i'm super super bad at these
\[\int\limits_{}^{}\frac{ (\tan^{-1}x)^{2} }{ 1+x^{2} }\]
is this like... replace 1+x^2 with cos or something?
what is the derivative of arctan(x)?
im sorry, had to verify email.
derivative of arctan = 1/1+x^2
do you see 1/(1+x^2) above?
\[\int\limits_{}^{}(\tan^{-1}(x))^2 \frac{1}{1+x^2}dx \] so if you replace arctan(x) with u then du=1/(1+x^2) dx
so what do you have?
(tan^-1x)^3/3?
+ c lol
ok cool
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