Question

Evaluate integral (e^x/sqrt(1+e^x+e^2x))dx
Trig substitution?

Answer 1

\[\int\limits_{}^{}\frac{e^x}{\sqrt{e^{2x}+e^x+\frac{1}{4}+1-\frac{1}{4}}} dx\]

\[\int\limits_{}^{}\frac{e^x}{\sqrt{(e^{x}+\frac{1}{2})^2+\frac{3}{4}}} dx\]

Answer 2

i would try trig substitution

Answer 3

\[\text{ let} \tan(\theta)=\frac{e^x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\]

Answer 4

did you complete the square?

Answer 5

yep

Answer 6

\[u^2+u+1=u^2+u+(\frac{1}{2})^2+1-(\frac{1}{2})^2=(u+\frac{1}{2})^2+\frac{3}{4}\]
if you don't like the exp(x) just replace it with u for now

Answer 7

Thank you
then i just substitute u right?

Answer 8

i would use the substitution i suggested above

Answer 9

\[\frac{\sqrt{3}}{2} \tan(\theta)=e^x+\frac{1}{2} => \frac{\sqrt{3}}{2} \tan(\theta)-\frac{1}{2}=e^x\]

\[\frac{\sqrt{3}}{2} \sec^2(\theta) d \theta=e^x dx\]

\[\int\limits_{}^{}\frac{\frac{\sqrt{3}}{2} \sec^2(\theta)}{ (\frac{\sqrt{3}}{2}\tan(\theta))^2+\frac{3}{4}} d \theta\]

Answer 10

oops i left the square root off

Answer 11

\[\int\limits\limits_{}^{}\frac{\frac{\sqrt{3}}{2} \sec^2(\theta)}{ \sqrt{(\frac{\sqrt{3}}{2}\tan(\theta))^2+\frac{3}{4}}} d \theta\]

Answer 12

\[\frac{\sqrt{3}}{2}\int\limits_{}^{}\frac{\sec^2(\theta)}{\sqrt{\frac{3}{4}\tan^2(\theta)+\frac{3}{4}}} d \theta\]

Answer 13

\[\frac{\frac{\sqrt{3}}{2}}{\sqrt{\frac{3}{4}}} \int\limits_{}^{}\frac{\sec^2(\theta) }{ \sqrt{\tan^2(\theta)+1}} d \theta\]

Answer 14

\[\int\limits_{}^{}\frac{\sec^2(\theta)}{\sec(\theta)} d \theta\]

Answer 15

\[\int\limits_{}^{}\sec(\theta)d \theta=\ln|\sec(\theta)+\tan(\theta)|+C\]

don't forget to put back in terms of x

Answer 16

Thank you

Answer 17

\[\tan \theta = e ^{x}+1/2/\sqrt{3}/2\]

Answer 18

what's \[\sec \theta\]

Answer 19

\[\frac{hyp}{adj}\]

Answer 20

yes, but in terms of x here