Ask your own question, for FREE!
Mathematics 11 Online
OpenStudy (anonymous):

[\int\limits_{}^{}\frac{ e^{4/x} }{ x^{2} }] please help :)

OpenStudy (anonymous):

\[\int\limits_{}^{}\frac{ e^{4/x} }{ x^{2} }\]

OpenStudy (amistre64):

let u = 4/x , what is dx?

OpenStudy (anonymous):

4lnx?

OpenStudy (amistre64):

no, derivative of: u = 4/x

OpenStudy (anonymous):

i know 1/x is lnx

OpenStudy (anonymous):

its not 4 ln x?

myininaya (myininaya):

he isn't asking you to integrate 1/x

OpenStudy (amistre64):

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?

OpenStudy (anonymous):

i understand, it's 4/x wouldn't it be 4lnx

OpenStudy (anonymous):

oh wow i was integrating

OpenStudy (amistre64):

\[\frac{e^{4/x}}{x^2}dx\] \[\frac{e^{u}}{x^2}dx\] we need to replace the other x parts with u parts letting u = 4/x

OpenStudy (amistre64):

latex seems to be on the fritx

OpenStudy (amistre64):

**fritz

OpenStudy (anonymous):

so it's deriv of 4x^-1 = -4x^-2

OpenStudy (amistre64):

almost du = -4x^(-2) dx -du/4 = x^(-2) dx

OpenStudy (anonymous):

ah i see now

OpenStudy (amistre64):

e^u ---- du is our substitution -4

OpenStudy (amistre64):

now, this is a constant and a rather simple e^u run up

OpenStudy (anonymous):

so (-1/4)(e^4/x)+c?

OpenStudy (amistre64):

take the derivative to chk

OpenStudy (amistre64):

http://www.wolframalpha.com/input/?i=derivative++-e%5E%284%2Fx%29%2F4 it works :)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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 :(

OpenStudy (anonymous):

my next questions are trig integrations i'm super super bad at these

OpenStudy (anonymous):

\[\int\limits_{}^{}\frac{ (\tan^{-1}x)^{2} }{ 1+x^{2} }\]

OpenStudy (anonymous):

is this like... replace 1+x^2 with cos or something?

myininaya (myininaya):

what is the derivative of arctan(x)?

OpenStudy (anonymous):

im sorry, had to verify email.

OpenStudy (anonymous):

derivative of arctan = 1/1+x^2

myininaya (myininaya):

do you see 1/(1+x^2) above?

myininaya (myininaya):

\[\int\limits_{}^{}(\tan^{-1}(x))^2 \frac{1}{1+x^2}dx \] so if you replace arctan(x) with u then du=1/(1+x^2) dx

myininaya (myininaya):

so what do you have?

OpenStudy (anonymous):

(tan^-1x)^3/3?

OpenStudy (anonymous):

+ c lol

myininaya (myininaya):

ok cool

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!