How do you integrate #h(t)=(t+1)/(t^2+2t+2)# using the quotient rule?

1 Answer
Nov 4, 2016

Please see the explanation for the steps leading to the answer:

#int(t + 1)/(t^2 + 2t + 2)dx = (1/2)ln(t^2 + 2t + 2) + C#

Explanation:

The quotient rule does not pertain to integrals; it pertains to derivatives.

This integral is best integrated by u substitution:

let #u = t^2 + 2t + 2#, then #du = (2t + 2)dx#, therefore, we shall substitute #(1/2)du# for #(t + 1)dx#:

#int(t + 1)/(t^2 + 2t + 2)dx = (1/2)int(du)/u#

#int(t + 1)/(t^2 + 2t + 2)dx = (1/2)ln|u| + C#

Reverse the substitution for u:

#int(t + 1)/(t^2 + 2t + 2)dx = (1/2)ln|t^2 + 2t + 2| + C#

Because #t^2 + 2t + 2# can never be less than or equal to zero, we may drop the absolute value:

#int(t + 1)/(t^2 + 2t + 2)dx = (1/2)ln(t^2 + 2t + 2) + C#