How do you integrate #int 1/sqrt(e^(2x)+12e^x+35)dx#?
1 Answer
# int \ 1/sqrt(e^(2x)+12e^x+35) \ dx = - 1/sqrt(35) \ ln|sqrt((35e^(-x)+6)^2-1)+35e^(-x)+6| + C #
Explanation:
We seek:
# I = int \ 1/sqrt(e^(2x)+12e^x+35) \ dx #
# \ \ = int \ 1/sqrt(e^(2x)(1+12e^(-x)+35e^(-2x))) \ dx #
# \ \ = int \ 1/(e^xsqrt((1+12e^(-x)+35e^(-2x)))) \ dx #
# \ \ = int \ (e^(-x))/(sqrt(1+12e^(-x)+35e^(-2x))) \ dx #
We can perform a substitution, Let:
# u = 35e^(-x)+6 => (du)/dx = -35e^(-x) # , and,#e^(-x)=(u-6)/35#
The we can write the integral as:
# I = int \ (-1/35)/(sqrt(1+12((u-6)/35)+35((u-6)/35)^2)) \ du #
# \ \ = -1/35 \ int \ (1)/(sqrt(1 + 12/35(u-6) + 1/35(u-6)^2)) \ du #
# \ \ = -1/35 \ int \ (1)/(sqrt(1/35(35 + 12(u-6) + (u-6)^2))) \ du #
# \ \ = -1/35 \ int \ (1)/(1/sqrt(35) \ sqrt(35 + 12u-72 + u^2-12u+36)) \ du #
# \ \ = -1/sqrt(35) \ int \ 1/sqrt(u^2-1) \ du #
This is a standard integral, so we can write:
# I = - 1/sqrt(35) \ ln|sqrt(u^2-1)+u| + C #
And if we restore the substitution, we get:
# I = - 1/sqrt(35) \ ln|sqrt((35e^(-x)+6)^2-1)+35e^(-x)+6| + C #
