Integrate e^2x/(16+e^4x) and sec^2(x)(tan(x)+6)^5

[brittini14]

1) Integrate e^2x/(16+e^4x)

2) Integrate sec^2(x)(tan(x)+6)^5

 

[tkhunny]

Please show your work. Give us a clue how to help you.

 

[Deleted member 4993]

1) Integrate e^2x/(16+e^4x)

substitute u = e^(2x)

2) Integrate sec^2(x)(tan(x)+6)^5

use d(tanx)/dx = sec^2x

 

[brittini14]

if you substitue u=e^2x you get du=2e^2x and then i don't know where to go from there .. here is the problem again

integrate (e^2x)/(16+e^4x)

 

[pka]

integrate (e^2x)/(16+e^4x)

This is in arctangent form. Look that up in your textbook.

 

[Deleted member 4993]

if you substitue u=e^2x you get du=2e^2x and then i don't know where to go from there .. here is the problem again

integrate (e^2x)/(16+e^4x)

u = e^(2x)

du = 2 e^(2x) dx

integrate (e^2x)/(16+e^4x) dx

= 1/2 * integrate du/(16+u^2)

= 1/2 * integrate du/(4^2 + u^2)


The above is a standard form - look up in your textbook

 

[brittini14]

thank you so much! that helped a lot! i got the arctan form from that!!