\( I = \int\frac{2x}{(x^2 + 1)(x^2 + 2)^2}dx\)
Let x2 = t
2x dx = dt
\(\therefore I = \int \frac{dt}{(t+1)(t+2)^2}\)
Let \(\frac{1}{(t + 1)(t + 2)^2} = \frac{A}{t+1}+ \frac{B}{t+2}+\frac{C}{(t+2)^2}\)
⇒ I = A (t + 2)2 + B(t +1) (t + 2) + c(t +1)
Put t = -1, we get
A = 1
Put t = -2, we get
C = -1
Put t = 0, we get
4A + 2B + C = 1
⇒ 2B + 4 - 1 = 1
⇒ 2B = -2
⇒ B = -1
\(\therefore \frac{1}{(t +1)(t +2)^2}= \frac1{t+1}-\frac1{t+2}-\frac1{(t+2)^2}\)
\(\therefore I = \int \left(\frac{1}{t +1}- \frac1{t+ 2}-\frac1{(t + 2)^2}\right)dt = log(t +1) - log (t +2) + \frac1{t+2}+C\)
\(= log \left|\frac{t+1}{t+2}\right|+ \frac1 {t+2}+C \)
\(= \log\left|\frac{x^2 +1}{x^2 +2 }\right|+\frac{1}{x^2+2}+C\)
