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- anonymous

I have question in substation rule

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- anonymous

I have question in substation rule

- schrodinger

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- anonymous

- anonymous

in substitution rule

- myininaya

all those are z's right?

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- anonymous

yep

- myininaya

did you let u=the bottom

- myininaya

\[u=e^z +z => du=(e^z+1 ) dz\]

- anonymous

yep i did in that way

- myininaya

\[\int\limits_{e^0+0}^{e^1+1}\frac{du}{u}\]

- myininaya

\[\ln|u||_1^{e^1+1}\]

- myininaya

\[\ln|e^1+1|-\ln|1|=\ln(e+1)-0=\ln(e+1)\]

- anonymous

yeah that is the right answer
Thank you so much :)

- myininaya

Do you understand all the steps I took?

- anonymous

i didn't understand one part of it which is the one before last step of it

- myininaya

you mean when I plugged in the upper limit and then minus plugged in the lower limit?

- anonymous

yeah
why did you put e to 1 + 1 and e to 0 +0?

- myininaya

so before the substitution the limits were 0

- anonymous

Oh I see
Thank you so much that was very helpful :)

- anonymous

I understood each step that you had done

- myininaya

Awesome! :)

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