How do you find the integral #ln( x + sqrt(x^2 -1)) #?

1 Answer
George C.
Mar 14, 2018

#int ln(x+sqrt(x^2-1)) dx = x ln(x+sqrt(x^2-1)) - sqrt(x^2-1) + C#

Explanation:

Use a hyperbolic substitution.

Put:

#x = cosh u#

Then:

#dx/(du) = sinh u#

and:

#u = ln(x+sqrt(x^2-1))#

and:

#int ln(x+sqrt(x^2-1)) dx = int ln(cosh u + sqrt(cosh^2 u - 1)) dx/(du) color(white)(.) du#

#color(white)(int ln(x+sqrt(x^2-1)) dx) = int ln(cosh u + sqrt(sinh^2 u)) sinh u color(white)(.) du#

#color(white)(int ln(x+sqrt(x^2-1)) dx) = int ln(cosh u + sinh u) sinh u color(white)(.) du#

#color(white)(int ln(x+sqrt(x^2-1)) dx) = int ln(e^u) sinh u color(white)(.) du#

#color(white)(int ln(x+sqrt(x^2-1)) dx) = int u sinh u color(white)(.) du#

#color(white)(int ln(x+sqrt(x^2-1)) dx) = int (u sinh u + cosh u)-cosh u color(white)(.) du#

#color(white)(int ln(x+sqrt(x^2-1)) dx) = u cosh u - sinh u + C#

#color(white)(int ln(x+sqrt(x^2-1)) dx) = x ln(x+sqrt(x^2-1)) - sqrt(x^2-1) + C#

Related questions
See all questions in Integration by Parts
Impact of this question
5774 views around the world
You can reuse this answer
Creative Commons License