Question

How do you solve the integral `int e^x/sqrt(e^(2x) + e^x+ 1)dx`?

Answer 1

I got `lnabs(2e^x+1+2sqrt(e^(2x)+x+1))+C`

Explanation:

We let `u = e^x`. Then `du = e^xdx` and `dx = (du)/e^x`.

`I = int u/(usqrt(u^2 + u + 1))`

`I = int1/sqrt(u^2 + u + 1)du`

`I = int 1/sqrt(1(u^2 + u + 1/4 - 1/4) + 1)du`

`I = int 1/sqrt((u + 1/2)^2 + 3/4) du`

We now let `t= (u +1/2)`. Then `dt = du`.

`I = int 1/sqrt(t^2 + 3/4) dt`

Now we let `t = sqrt(3)/2tantheta`. Then `dt = sqrt(3)/2sec^2theta`

`I = int 1/sqrt((sqrt3/2tantheta)^2 + 3/4) * sqrt(3)/2sec^2theta d theta`

`I = int 1/sqrt(3/4tan^2theta + 3/4) * sqrt(3)/2sec^2theta d theta`

Now we use `tan^2theta + 1 = sec^2theta`.

`I = int 1/sqrt(3/4sec^2theta) * sqrt(3)/2sec^2theta d theta`

`I = int 1/(sqrt(3)/2sectheta) * sqrt(3)/2sec^2theta d theta`

`I = int2/sqrt(3)costheta * sqrt(3)/2sec^2theta d theta`

We know that cosine and secant are reciprocals, so their product is `1`.

`I = int secthetad theta`

This is a common integral, derived here

`I = ln|sectheta + tantheta| + C`

We know from our initial substitution that `tantheta = (2t)/sqrt(3)`, and because `tan^2theta + 1 = sec^2theta`, we know that

`sec^2theta = (4t^2)/3 + 1 = (4t^2 + 3)/3`

And

`sectheta = sqrt((4t^2 + 3)/3)`

Thus,

`I = ln|(2t)/sqrt(3) + sqrt((4t^2 + 3)/3)| + C`

Now we can reverse the others substitutions.

`I = ln|(2(u+ 1/2))/sqrt(3) + sqrt((4(u + 1/2)^2 + 3)/3)| +C`

`I = ln|(2(e^x + 1/2))/sqrt(3) + sqrt((4(e^x + 1/2)^2 + 3)/3)| + C`

Note that `sqrt(4(e^x+1/2)^2+3)=sqrt(4(e^(2x)+e^x+1/4)+3)=sqrt(4e^(2x)+4e^x+4)=2sqrt(e^(2x)+e^x+1)`

`I=lnabs((2e^x+1+2sqrt(e^(2x)+x+1))/(sqrt3))+C`

Note that `sqrt3` can be removed from the natural logarithm as `-lnsqrt3`, which will be absorbed into the constant of integration `C`. This leaves:

`I=lnabs(2e^x+1+2sqrt(e^(2x)+x+1))+C`

Hopefully this helps!