Verified answer

* nghàm dx/(x²+a²)

đặt x = a.tant => dx = a(1+tan²t)dt

I = nghàm a(1+tan²t)dt /(a²tan²t + a²) = (1/a).nghàm(dt) = (1/a).t + C

* nghàm dx/(x²-a²) phân tích ra:

1/(x²-a²) = (1/2a).[1/(x-a) - 1/(x+a)]

nghàm... = (1/2a).[ln|x-a| - ln|x+a|] + C = (1/2a).ln|(x-a)/(x+a)| + C

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*nguyen ham cua dx/(x^2+a^2) :

dat x=!a!tant ( voi t thuoc (-pi/2;pi/2) )

suy ra dx=!a!.1/cos^2(t) .dt = !a!.(1+tan^2(t)) .dt

hoac co the dat: x=!a!cot(t) ( voi t thuoc (0;pi) )

suy ra dx=-!a!.1/sin^2(t) .dt = -!a!.(1+cot^2(t)).dt

*nguyen ham cua dx/(x^2-a^2)

dat x=!a!/sint ( voi t thuoc [-pi/2;pi/2]\{0} )