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I had a lecture earlier today where the use of partial fractions was introduced. He used partial fractions and a more 'brute force' method to $\int\frac{1}{(x^2 + 5x + 6)}\mathrm dx$. I could solve this using partial fractions but I need to be reminded of the more difficult method(which I've learned months ago) for my current maths subject's purposes. I've been trying to find a solution that yields to $\ln\left|\frac{(x + 2)}{(x + 3)}\right| + C$ to no avail. Can anyone help me how to solve the problem without using partial fractions?

abiessu
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Tristran Thorn
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    Partial fractions are very important as an algebra technique, not just for integration. Did you try turning the expression into $q^2+k$ for $q=ax+b$ and $k$ a constant? – abiessu Aug 14 '15 at 13:47

2 Answers2

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By completing the square,

$$x^2+5x+6=\frac{(2x+5)^2-1}4.$$

Then use the change of variable $t=2x+5$ and get

$$I=\int\frac{dx}{x^2+5x+6}=2\int\frac{dt}{t^2-1}.$$

You should recognize the derivative of the inverse hyperbolic tangent

$$I=-2\,\text{artanh}(t)=\ln\left(\left|\frac{1-t}{1+t}\right|\right)=\ln\left(\left|\frac{x+2}{x+3}\right|\right).$$

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$\dfrac 1 {x^2+5x+6}=\dfrac 1 {(x+3)(x+2)}=\dfrac 1 {x+2}-\dfrac 1 {x+3} $ $\int \dfrac {dx} {x^2+5x+6}=\int \dfrac {dx} {x+2} -\int \dfrac {dx} {x+3} =ln|x+2|-ln|x+3| +c=ln|\dfrac {x+2} {x+3} |+c$

Booldy
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