Need help with integral [0 to 1] of [(e^x) + 1] / [(e^x) + 2] dx

[ricecrispie]

Hi everyone,

We have to find the definite integral of:

[(e^x) + 1] / [(e^x) + 2] dx

The upper limit being 1 and the lower being 0.

I want to try it out myself however I can't seem to find an appropiate expressiom of u = (substitution method)

Please can someone hint on what to do

Sent from my LG-H840 using Tapatalk

 

[Dr.Peterson]

Hi everyone,

We have to find the definite integral of:

[(e^x) + 1] / [(e^x) + 2] dx

The upper limit being 1 and the lower being 0.

I want to try it out myself however I can't seem to find an appropiate expressiom of u = (substitution method)

Please can someone hint on what to do

Sent from my LG-H840 using Tapatalk

I would take u=e^x. The substitution may work a little differently from most you have done, but it will be valid. Please show your work if you have trouble.

 

[ricecrispie]

I would take u=e^x. The substitution may work a little differently from most you have done, but it will be valid. Please show your work if you have trouble.

I had been writing the question down wrong, I know what to do now thanks!

Sent from my LG-H840 using Tapatalk

 

[HallsofIvy]

I would first do the division. \(\displaystyle \frac{e^x+ 1}{e^x+ 2}= 1- \frac{1}{e^x+ 2}\). Now let \(\displaystyle u= e^x+ 2\) so that \(\displaystyle du= e^xdx\), \(\displaystyle dx= \frac{1}{e^x}du= \frac{1}{u- 2}du\).


\(\displaystyle \int \frac{e^x+ 1}{e^x+ 2}= \int dx - \int\frac{1}{u(u- 2)}du\).