Integral of x/(x + 1)²(x+2)²

Question 1

Question 2

I suppose you mean the integral

$\displaystyle \int (x)/((x+1)^2 (x+2)^2) \, dx$

Break up the integrand into partial fractions:

(x)/((x+1)^2 (x+2)^2) = (a)/(x+1) + (b)/((x+1)^2) + (c)/(x+2) + (d)/((x+2)^2)

$(x)/((x+1)^2 (x+2)^2) \\ = (a(x+1)(x+2)^2 + b(x+2)^2 + c(x+1)^2(x+2) + d(x+1)^2)/((x+1)^2 (x+2)^2)$

$(x)/((x+1)^2 (x+2)^2) \\ = ((a+c)x^3+(5a+b+4c+d)x^2+(8a+4b+5c+2d)x + 4a+4b+2c+d)/((x+1)^2(x+2)^2)$

x = (a+c)x^3+(5a+b+4c+d)x^2+(8a+4b+5c+2d)x + 4a+4b+2c+d

Solve for the coefficients:

$\begin{cases}a+c=0 \\ 5a+b+4c+d=0 \\ 8a+4b+5c+2d=1 \\ 4a+4b+2c+d=0\end{cases} \implies a=3,b=-1,c=-3,d=-2$

So we have

$\displaystyle \int (x)/((x+1)^2 (x+2)^2) \, dx= \int \left(\frac3{x+1} - \frac1{(x+1)^2} - \frac3{x+2} - \frac2{(x+2)^2}\right) \, dx$

$\displaystyle \int (x)/((x+1)^2 (x+2)^2) \, dx= 3\ln|x+1| + \frac1{x+1} - 3\ln|x+2| + \frac2{x+2} + C$

$\displaystyle \int (x)/((x+1)^2 (x+2)^2) \, dx= \ln|x+1|^3 - \ln|x+2|^3 + (3x+4)/((x+1)(x+2)) + C$

$\displaystyle \int (x)/((x+1)^2 (x+2)^2) \, dx= \boxed(x+1)/(x+2)\right$

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