# Find this integral in terms of the given integrals

1. Oct 10, 2013

### utkarshakash

1. The problem statement, all variables and given/known data
If $\displaystyle \int_0^1 \dfrac{e^t}{t+1} dt = a$ then $\displaystyle \int_{b-1}^b \dfrac{e^{-t}}{t-b-1} dt$ is equal to

2. Relevant equations

3. The attempt at a solution

I used the definite integral property in the second integral

$\displaystyle \int_{b-1}^b \dfrac{e^{-2b-1-t}}{b-2-t} dt$

2. Oct 10, 2013

### brmath

They want you to solve the second integral in terms of the a they gave you.

Try making the substitution u = t - b + 1 in the second integral.

3. Oct 11, 2013

### utkarshakash

Thanks. But how did it occur to you that u=t-b+1 would be a good idea?

4. Oct 12, 2013

### Saitama

You need to change the limits to 0 and 1. So our aim is to find a substitution that would do this job.

5. Oct 12, 2013

### brmath

Well, I was trying to change your second integral so it would look more like the first. In general it is useful to get rid of the constants in the denominator. And as Pranav-Arora points out that formula is handy for getting the limits of integration right.

The problem now is that with the change u = t - b + 1 you don't have the right denominator. And if you set up u to get u + 1 in the denominator you don't have the right limits of integration.

So I'm not sure this trick is the right approach. I'll think about it some more.

6. Oct 12, 2013

### Saitama

Your substitution is OK, you need to play with the given integral.

$$\int_0^1 \dfrac{e^t}{t+1} dt = \int_0^1 \dfrac{e^{1-t}}{2-t} dt$$

Do you see now?

7. Oct 12, 2013

### brmath

Yes, after all that I made the wrong substitution in the e$^{-t}$. Thank you for your help.

Last edited: Oct 12, 2013