How do you use substitution to integrate #(2x+7)/(x^2+5x+6) dx#?

Question

Charles Anctil · Accepted Answer

The answer can be written as #int (2x+7)/(x^2+5x+6)\ dx=ln|((x+2)^3)/(x+3)|+C#

You don't need substitution to do this integral. Use the Method of Partial Fractions instead.

Set # (2x+7)/(x^2+5x+6)=A/(x+2)+B/(x+3)# and solve for #A# and #B# (note that #x^2+5x+6=(x+2)(x+3)#).

Multiply both sides of the preceding equation by #(x+2)(x+3)# to get, after cancellation, #2x+7=A(x+3)+B(x+2)#. The quickest way to solve for #A# and #B# is to substitute #x=-3# and #x=-2# into this last equation (even though they make the original equation undefined).

#x=-3\Rightarrow 1=-B\Rightarrow B=-1#

#x=-2\Rightarrow 3=A#

Hence, # (2x+7)/(x^2+5x+6)=3/(x+2)-1/(x+3)#

Now substitute, integrate, and use properties of logarithms:

#int (2x+7)/(x^2+5x+6)\ dx=int (3/(x+2)-1/(x+3))\ dx#

#=3ln|x+2|-ln|x+3|+C=ln|((x+2)^3)/(x+3)|+C#

Cameron Alexander · Answer

undefined To integrate the given rational function using substitution, you can first decompose the denominator into linear factors, then proceed with a partial fraction decomposition. This involves expressing the rational function as a sum of simpler fractions. Once you have the partial fraction decomposition, you can integrate each term individually. After integrating, you can then substitute back to obtain the final result.