Question
Find: $\int \frac{3 x-2}{(x+1)^{2}(x+3)} d x$
✓
Answer
We write
$\frac{3 x-2}{(x+1)^{2}(x+3)}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C}{x+3}$
So that 3x - 2 = A (x + 1) (x + 3) + B (x + 3) + C (x + 1)2
= A (x2 + 4x + 3) + B (x + 3) + C (x2 + 2x + 1)
Comparing coefficient of x2, x and constant term on both sides,
we get A + C = 0, 4A + B + 2C = 3 and 3A + 3B + C = -2. Solving these equations, we get
$A=\frac{11}{4}, B=\frac{-5}{2}$ and $C=\frac{-11}{4}$. Thus the integrand is given by
$\frac{3 x-2}{(x+1)^{2}(x+3)}=\frac{11}{4(x+1)}-\frac{5}{2(x+1)^{2}}-\frac{11}{4(x+3)}$
Therefore, $\int \frac{3 x-2}{(x+1)^{2}(x+3)}=\frac{11}{4} \int \frac{d x}{x+1}-\frac{5}{2} \int \frac{d x}{(x+1)^{2}}-\frac{11}{4} \int \frac{d x}{x+3}$
= $\frac{11}{4} \log |x+1|+\frac{5}{2(x+1)}-\frac{11}{4} \log |x+3|+\mathrm{C}$
= $\frac{11}{4} \log \left|\frac{x+1}{x+3}\right|+\frac{5}{2(x+1)}+C$
$\frac{3 x-2}{(x+1)^{2}(x+3)}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C}{x+3}$
So that 3x - 2 = A (x + 1) (x + 3) + B (x + 3) + C (x + 1)2
= A (x2 + 4x + 3) + B (x + 3) + C (x2 + 2x + 1)
Comparing coefficient of x2, x and constant term on both sides,
we get A + C = 0, 4A + B + 2C = 3 and 3A + 3B + C = -2. Solving these equations, we get
$A=\frac{11}{4}, B=\frac{-5}{2}$ and $C=\frac{-11}{4}$. Thus the integrand is given by
$\frac{3 x-2}{(x+1)^{2}(x+3)}=\frac{11}{4(x+1)}-\frac{5}{2(x+1)^{2}}-\frac{11}{4(x+3)}$
Therefore, $\int \frac{3 x-2}{(x+1)^{2}(x+3)}=\frac{11}{4} \int \frac{d x}{x+1}-\frac{5}{2} \int \frac{d x}{(x+1)^{2}}-\frac{11}{4} \int \frac{d x}{x+3}$
= $\frac{11}{4} \log |x+1|+\frac{5}{2(x+1)}-\frac{11}{4} \log |x+3|+\mathrm{C}$
= $\frac{11}{4} \log \left|\frac{x+1}{x+3}\right|+\frac{5}{2(x+1)}+C$
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