Integrate the int 1/(x^2-6x+11) dx from -5 to 5?
1 Answer
Jun 19, 2018
Explanation:
Let
#I=int_-5^5 1/(x^2-6x+11)dx#
Complete the square in the denominator:
#I=int_-5^5 1/((x-3)^2+2)dx#
Apply the substitution
#I=1/sqrt2intd theta#
Integrate directly:
#I=1/sqrt2[theta]#
Reverse the substitution:
#I=1/sqrt2[tan^(-1)((x-3)/sqrt2)]_-5^5#
Insert the limits of integration:
#I=1/sqrt2(tan^(-1)(sqrt2)+tan^(-1)(4sqrt2))#