Why is the indefinite integral of et(i - 1)/(i - 1) + C not equal to et(i - 1)?

[Miike012]

The integral is in the attachment.

Why is the indefinite integral not equal to...
e^{t(i - 1)}/(i - 1) + C

Because d/dt[e^{t(i - 1)}/(i - 1) + C] = e^{t(i - 1)}

 

[Curious3141]

The integral is in the attachment.

Why is the indefinite integral not equal to...
e^{t(i - 1)}/(i - 1) + C

Because d/dt[e^{t(i - 1)}/(i - 1) + C] = e^{t(i - 1)}

If $i$ represents the square root of minus 1, then your answer is right, and the integral in the attachment is wrong. Where did you get that from?

But you should put $\frac{1}{1-i}$ in the form of $a+bi$ in the final answer.

 

[jeppetrost]

It seems that what ever calculated that didn't identify the t in the exp(...) and the t in dt. It is either bad notation or plain wrong.

 

[clamtrox]

Don't trust wolframalpha blindly. It has some funny ways of guessing what you mean, and in this case it guesses completely wrong.

 

[jeppetrost]

I was indeed able to reproduce the (wrong) result on WA. What you've done is *not* put an asterisk between t and (i-1) in the exp(). This means that WA thinks the t in the exp is maybe some kind of function t(x) [evaluated at x=i-1] and you're integrating wrt. some other variable t.
In short, if you write "int exp(t*(i-1)),t" and "int exp(t(i-1)),t" you get different results.