How do you integrate #int 1/sqrt(e^(2x)+12e^x+35)dx#?

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Steve M Share
May 18, 2018

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 #

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