Let's find the indefinite integral of \(\dfrac{e^x}{x}.\)

....I know it's \(\text{Ei}(x)+C\) but let's do it anyways

So firstly, I'm gonna let \(\dfrac{1}{x}=t,\) leading us to \(x=\dfrac{1}{t}.\)

Differentiating gives \(dx=-\dfrac{1}{t^2}dt,\) and therefore:

\[\begin{align} \int \frac{e^x}{x}dx &= \int \frac{1}{x}\cdot e^x dx \\ & =\int t \cdot e^{\frac{1}{t}}\cdot\left(-\dfrac{1}{t^2}\right) dt \\ & =- \int \frac{e^{\frac{1}{t}}}{t} dt \\ & =- \int \frac{e^{\frac{1}{x}}}{x} dx \\ \end{align}\]

From above, we get:

\(\displaystyle \int \frac{e^x}{x} dx + \int \frac{e^{\frac{1}{x}}}{x} dx = 0 + (C)\)

\(\displaystyle \int \frac{e^x+e^{\frac{1}{x}}}{x} dx = 0 + (C)\)

Differentiating both sides gives:

\(\displaystyle \frac{e^x + e^{\frac{1}{x}}}{x} = 0\)

....then...

\(\displaystyle e^x+e^{\frac{1}{x}}=0\)...?!

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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TopNewest\(\int \frac{e^{1/t}}{t} dt \neq \int \frac{e^{1/x}}{x} dx\). Remember \(x=\frac{1}{t}\), 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 \(\displaystyle \int \frac{e^x}{x}dx+\int \frac{e^{\frac{1}{x}}}{x}dx=C,\) but when you

differentiateboth sides,\(\dfrac{e^x+e^{\frac{1}{x}}}{x}=0.\)

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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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