Question

Integral of x"2+4/x"2+4x+3

Answer 1

I'm guessing you mean

`\displaystyle \int(x^2+4)/(x^2+4x+3)\,\mathrm dx`

First, compute the quotient:

`\displaystyle (x^2+4)/(x^2+4x+3) = 1 + (4x-1)/(x^2+4x+3)`

Split up the remainder term into partial fractions. Notice that

x ² + 4x + 3 = (x + 3) (x + 1)

Then

`\displaystyle (4x-1)/(x^2+4x+3) = \frac a{x+3} + \frac b{x+1} \\\\ \implies 4x - 1 = a(x+1) + b(x+3) = (a+b)x + a+3b \\\\ \implies a+b=4 \text{ and }a+3b = -1 \\\\ \implies a=\frac{13}2\text{ and }b=-\frac52`

So the integral becomes

`\displaystyle \int \left(1 + (13)/(2(x+3)) - (5)/(2(x+1))\right) \,\mathrm dx = \boxed{x + \frac{13}2\ln|x+3| - \frac52 \ln|x+1| + C}`

We can simplify the result somewhat:

`\displaystyle x + \frac{13}2\ln|x+3| - \frac52 \ln|x+1| + C \\\\ = x + \frac12 \left(13\ln|x+3| - 5\ln|x+1|\right) + C \\\\ = x + \frac12 \left(\ln\left|(x+3)^(13)\right| - \ln\left|(x+1)^5\right|\right) + C \\\\ = x + \frac12 \ln\left|((x+3)^(13))/((x+1)^5)\right| + C \\\\ = \boxed{x + \ln\sqrt\left + C}`