How do you integrate #int 1/sqrt(-e^(2x)-12e^x-37)dx#?

1 Answer
May 10, 2018

Answer:

# int \ 1/sqrt(-e^(2x)-12e^x-37) \ dx = i/sqrt(37) \ "arcsinh" \ (37e^(-x)+6) + C #

Explanation:

We seek:

# I = int \ 1/sqrt(-e^(2x)-12e^x-37) \ dx #

# \ \ = int \ 1/sqrt((-1)e^(2x)(1+12e^(-x)+37e^(-2x))) \ dx #

# \ \ = int \ 1/(ie^xsqrt((1+12e^(-x)+37e^(-2x)))) \ dx #

# \ \ = int \ (-ie^(-x))/(sqrt(1+12e^(-x)+37e^(-2x))) \ dx #

We can perform a substitution, Let:

# u = 37e^(-x)+6 => (du)/dx = -37e^(-x) #, and, #e^(-x)=(u-6)/37#

The we can write the integral as:

# I = int \ (-i/37)/(sqrt(1+12((u-6)/37)+37((u-6)/37)^2)) \ du #

# \ \ = -i/37 \ int \ (1)/(sqrt(1 + 12/37(u-6) + 1/37(u-6)^2)) \ du #

# \ \ = -i/37 \ int \ (1)/(sqrt(1/37(37 + 12(u-6) + (u-6)^2))) \ du #

# \ \ = -i/37 \ int \ (1)/(1/sqrt(37) \ sqrt(37 + 12u-72 + u^2-12u+36)) \ du #

# \ \ = -i/sqrt(37) \ int \ 1/sqrt(u^2+1) \ du #

This is a standard integral, so we can write (temporarily omitting the constant of integration):

# I = -i/sqrt(37) \ "arcsinh" \ u #

And if we restore the substitution, we get:

# I = -i/sqrt(37) \ "arcsinh" \ (37e^(-x)+6) + C #