Let's find the indefinite integral of xex.
....I know it's Ei(x)+C but let's do it anyways
So firstly, I'm gonna let x1=t, leading us to x=t1.
Differentiating gives dx=−t21dt, and therefore:
∫xexdx=∫x1⋅exdx=∫t⋅et1⋅(−t21)dt=−∫tet1dt=−∫xex1dx
From above, we get:
∫xexdx+∫xex1dx=0+(C)
∫xex+ex1dx=0+(C)
Differentiating both sides gives:
xex+ex1=0
....then...
ex+ex1=0...?!
I didn't expect this to happen but I have this terrible migraine to think more deeply about this ._.
Yes I am stupid, sorry ;;>_>
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Top Newest∫te1/tdt=∫xe1/xdx. Remember x=t1, and we are talking about indefinite integrals.
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Yeah... Someone told me that in the below chain of comments and...
....I feel even dumber now. x'D
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You forgot about the arbitrary constant of integration.
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Hmm...
then let ∫xexdx+∫xex1dx=C, but when you differentiate both sides,
xex+ex1=0.
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[ this comment is wrong ]
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The integral of 1 - cos^2 x is
The integral of 1 - the integral of cos^2 x
= x - the integral of cos^2 x + C
Therefore
The integral of sin^2 x + the integral of cos^2 x = x + C
Differentiate and we get
sin^2 x + cos^2 x = 1
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Summoning the great @Mark Hennings
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F(x)=∫1xueuduG(x)=∫1xueu1du Then the substitution v=u−1 gives F(x)=−∫1x−1vev1dv=−G(x−1) and hence F(x)+G(x−1)=0 What you are saying is that F(x)+G(x)=c which is not true. You are making the substitution t=x−1, and then replacing t by x, instead of by x−1. Since you are working with indefinite integrals, you cannot exchange the variables freely if they already have an interrelationship.
Suppose that we writeLog in to reply
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