How do you integrate #int 1/sqrt(x^2+2x)# by trigonometric substitution?
1 Answer
Explanation:
Complete the square in the denominator (within the
#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
#int 1/sqrt(u^2 - 1)du#
Now use the substitution
#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
#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!