# Integrating $\int^{e^3-1}_{0}\frac{dt}{1+t}.$

How can I integrate $$\int^{e^3-1}_{0}\frac{dt}{1+t}.$$

I tried to make $u=1+t$ which means that $du=dt$ but it's not giving me anything useful, but instead made things more complicated. Maybe I did something wrong, but can someone tell me the correct way of solving this, or the correct $u$-substitution?

Thanks!

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Bear in mind that when $t=0$ then $u=1$ and when $t=e^3-1$ then $u=e^3$, so you get $\displaystyle\int_1^{e^3}$. And you should remember what $\ln(e^3)$ is; if you leave an answer as "$\ln(e^3)$" without simplifying then you're missing something important. –  Michael Hardy Jul 5 '13 at 18:44

Hint:

Use the fact that $\dfrac d{du}\left(\ln u\right) = \dfrac {u'}u:\;$ And so we have , $$\int \frac{du}{u} = \ln|u| + c$$

$$\int\frac{dt}{1+t}$$ Correctly, you let $u = 1 + t,\quad \,du = dt$.

This gives us $$\int \frac{du}{u}$$

I trust you can take it from here?!

Note: You can either

• change the bounds of integration by replacing the lower bound with $u$ evaluated at $x = 0$ and replacing the upper bound with $u$ evaluated at $x = e^3 - 1$, thereby keeping all subsequent work in terms of $u$, or
• you can integrate (as you would an indefinite integral) with respect to $u$, back-substitute by replacing $u$ in the result with $1 + t$, and use then evaluate that at the original bounds.
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Thanks for the hint! My instructor didn't teach that inclass or maybe we didn't get to that point, I was just trying to prep things ahead –  user85130 Jul 5 '13 at 18:37
You're all set then: Can you take it from here? –  amWhy Jul 5 '13 at 18:39
Yes, thanks for the help! –  user85130 Jul 5 '13 at 18:41
Why not just say $\ln(e^3)=3$ instead of recalling a value of $\ln e$? It's basic to what logarithms are that $\log_b(b^x) = x$. ${}\qquad{}$ –  Michael Hardy Jul 5 '13 at 18:45

We know that $dt=d(t+1)$ so you have an indefinite integral as follows first:

$$\int\frac{d(t+1)}{t+1}=\int\frac{du}{u}$$

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$\quad \tiny + \quad$ –  amWhy Jul 6 '13 at 1:38

$$\int\frac{dt}{1+t}=\int\frac{d(t+1)}{1+t}=\ln|1+t|+C$$

$$\implies \int^{e^3-1}_{0}\frac{dt}{1+t}=(\ln|1+t|+C)_0^{e^3-1}=\ln |e^3-1+1|-\ln|1+0|=\ln(e^3)=3\ln e=3$$

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@ThomasE., as $d(t+1)=dt$ –  lab bhattacharjee Jul 5 '13 at 18:39