Question

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

Answer 1

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

Explanation:

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`