How do you integrate #int 1/sqrt(x^2+2x)# by trigonometric substitution?

1 Answer
Noah G
Feb 20, 2017

#ln|sqrt((x + 1)^2 - 1) + x + 1| + C#

Explanation:

Complete the square in the denominator (within the #sqrt#).

#int 1/sqrt(1(x^2 + 2x + 1 - 1))dx#

#int 1/sqrt(1(x^2 + 2x + 1) - 1)dx#

#int 1/sqrt((x + 1)^2 - 1)dx#

Let #u = x + 1#. Then #du = dx#.

#int 1/sqrt(u^2 - 1)du#

Now use the substitution #u = sectheta#. Then #du = secthetatanthetad theta#.

#int 1/sqrt(sec^2theta - 1) secthetatantheta d theta#

#int 1/sqrt(tan^2theta) secthetatantheta d theta#

#int 1/tantheta secthetatantheta d theta#

#int sectheta d theta#

This is a known integral that can be derived here

#ln|sectheta + tantheta| + C#

Obviously it's not good enough to stay in #theta#; we have to return to #x#. From our initial substitution, #u/1 = sectheta#. This means that the side opposite #theta# measures #sqrt(u^2 - 1)#. This also means that #tantheta = sqrt(u^2 - 1)/1 = sqrt(u^2 - 1)#.

#ln|sqrt(u^2 - 1) + u| + C#

We have one more substitution to reverse.

#ln|sqrt((x + 1)^2 - 1) + x + 1| + C#

Hopefully this helps!

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