Integrate the int 1/(x^2-6x+11) dx from -5 to 5?

Question

1565 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-19T14:44:24

#int_-5^5 1/(x^2-6x+11)dx=1/sqrt2(tan^(-1)(sqrt2)+tan^(-1)(4sqrt2))#

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 #x-3=sqrt2tantheta#:

#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))#