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

Question

i can't seem to find the answere.. i started with subst e^x=t or e^(4x)=(e^(2x))^2 and subst e^2x=t but didnt get me anywhere
thx in advance

Answer 1

I'm assuming you mean: [e^(4x)+e^x]/[e^(4x)+e^(2x)]



Let u = e^x



[e^(4x)+e^x]/[e^(4x)+e^(2x)] = (u^4 + u)/(u^4 + u^2)

[e^(4x)+e^x]/[e^(4x)+e^(2x)] = (u^4 + u^2 - u^2 + u)/(u^4 + u^2)

[e^(4x)+e^x]/[e^(4x)+e^(2x)] = 1 - (u^2 - u)/(u^4 + u^2)

[e^(4x)+e^x]/[e^(4x)+e^(2x)] = 1 - (u - 1)/(u^3 + u)



Lets use partial fractions since we can factor the denominator.



(u - 1)/(u^3 + u) = A/u + (Bu + C)/(u^2 + 1)

u - 1 = A(u^2 + 1) + (Bu + C)u

u - 1 = Au^2 + A + Bu^2 + Cu

u - 1 = (A + B)u^2 + Cu + A



Equating coefficients we find A = -1, C = 1 and B = 1 (since A + B = 0)



(u - 1)/(u^3 + u) = -1/u + (u + 1)/(u^2 + 1)





[e^(4x)+e^x]/[e^(4x)+e^(2x)] = 1 - [-1/u + (u + 1)/(u^2 + 1)]

[e^(4x)+e^x]/[e^(4x)+e^(2x)] = 1 + 1/u - (u + 1)/(u^2 + 1)

[e^(4x)+e^x]/[e^(4x)+e^(2x)] = 1 + 1/u - u/(u^2 + 1) - 1/(u^2 + 1)



Integrating this with respect to u



∫[e^(4x)+e^x]/[e^(4x)+e^(2x)]dx = ∫[1 + 1/u - u/(u^2 + 1) - 1/(u^2 + 1)]dx



Now we have to determine dx in terms of u and du



u = e^x --> du = e^xdx = u dx --> dx = du/u



∫[e^(4x)+e^x]/[e^(4x)+e^(2x)]dx = ∫[1 + 1/u - u/(u^2 + 1) - 1/(u^2 + 1)](du/u)



∫[e^(4x)+e^x]/[e^(4x)+e^(2x)]dx = ∫[1/u + 1/u^2 - 1/(u^2 + 1) - 1/[u(u^2 + 1)]du



We need to use partial fractions again for the last term on the RHS.



1/[u(u^2 + 1)] = A/u + (Bu + C)/(u^2 + 1)

1 = A(u^2 + 1) + (Bu + C)u

1 = Au^2 + A + Bu^2 + Cu

1 = (A+B)u^2 + Cu + A



Equating the coefficients we find A=1, C=0 and B=-1 since A+B=0



1/[u(u^2 + 1)] = 1/u - u/(u^2 + 1)



Substitute this back into the integral.



∫[e^(4x)+e^x]/[e^(4x)+e^(2x)]dx = ∫[1/u + 1/u^2 - 1/(u^2 + 1) - 1/u + u/(u^2 + 1)]du



∫[e^(4x)+e^x]/[e^(4x)+e^(2x)]dx = ∫[1/u^2 - 1/(u^2 + 1) + u/(u^2 + 1)]du



Now the integral is solvable.



∫[e^(4x)+e^x]/[e^(4x)+e^(2x)]dx = ∫[1/u^2 - 1/(u^2 + 1) + u/(u^2 + 1)]du



∫[e^(4x)+e^x]/[e^(4x)+e^(2x)] = -1/u - arctan(u) + ½ln(u^2 + 1) + C



Substituting u = e^x yields



∫[e^(4x)+e^x]/[e^(4x)+e^(2x)] = -e^-x - arctan(e^x) + ½ln[e^(2x) + 1] + C



I am pretty sure this is correct.

Answer 2

Note: Your numerator has a parenthesis; your denominator doesn't.

Divide (e^(4x)+e^x) / e^(4x) first

= 1 + e^(-3x)

Integrating, x+e^(-3x) / (-3)

= x -e^(-3x) / 3 ----(1)

Now integrate e^(2x) = e^(2x)/2 ---(2)

(1)+(2) =

x - e^(-3x) / 3 + e^(2x) / 2 + C

Answer 3

if you go to www.wolfram.com



the integrator there can't do it without a magic function in the result, so unless that's wrong or you are in to magic functions..