Lời giải:
Đặt $x=tgx=>dx=(1+tg^2t)dt$.Khi đó:
$I=$$\int\limits {\frac{dx}{(1+tg^2t)^2}} \, =$ $\int\limits {cos^4tdt} \, =\frac{1}{4}$ $\int\limits {(1+cos2t)^2dt} \, =\frac{1}{4}$ $\int\limits {dt} \, +\frac{1}{2}$ $\int\limits {cos2tdt} \,+\frac{1}{4} $ $\int\limits {cos^22tdt} \,=\frac{1}{4}t+\frac{1}{4}sin2t+\frac{1}{8}$ $\int\limits {(1+cos4t)dt} \,=\frac{3}{8}t+\frac{1}{4}sin2t+\frac{1}{32}sin4t+C $
Trong đó:
$sin2t=\frac{2x}{1+x^2}$
$sin4t=\frac{4x.(1-x^2)}{(1+x^2)^2}$
=>$I=\frac{3}{8}arctgx+\frac{x}{2.(1+x^2)}+\frac{x.(1-x^2)}{8.(1+x^2)^2}+C$
