What is #int x/ sqrt(x^2 - 8^2) dx#?

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Answer 1 · 2016-02-04T13:12:04

# int x/(sqrt(x^2 - 64)) "d"x = sqrt(x^2 - 64) + C#

where #C# is a constant.

#int x/(sqrt(x^2 - 8^2)) "d"x = int x/(sqrt(x^2 - 64)) "d"x#

Let #u = sqrt(x^2 - 64)#. We need to differentiate #u# as next:

#("d"u)/("d"x) = 1 / (2 sqrt(x^2 - 64)) * 2x = (cancel(2)x)/(cancel(2) sqrt(x^2 - 64)) = x / (sqrt(x^2 - 64))#

Multiply by #"d"x# to have just #"d"u# on one side:

#"d"u = x / (sqrt(x^2 - 64)) "d"x#

As you can see, the substition seems to fit our integral perfectly. So we can start to substitute and solve the integral:

#int x/(sqrt(x^2 - 64)) "d"x = int 1 * color(blue)(x/(sqrt(x^2 - 64)) "d"x)#

... replace #x / (sqrt(x^2 - 64)) "d"x# by #"d"u# and all other occurences of #sqrt(x^2 - 64)#, if present, by #u#...

#= int 1 color(white)(ii) color(blue)("d"u)#

... the integral of #1# is #u + C# with a constant #C#...

#= u + C #

... re-substitute by replacing #u# with #sqrt(x^2 - 64)# ...

#= sqrt(x^2 - 64) + C#

Hope that this helped!