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求不定积分,用换元法!
1)∫1/根号(x^2+1)^3 dx
2)∫1/根号x+立方根号x dx
人气:335 ℃ 时间:2020-02-04 06:49:30
解答
1) 令:x=tant ,√(x^2+1)^3 = sec³t ,cost = 1/√(x^2+1) ,dx = sec²t dt ∫1/√(x^2+1)^3 dx=∫1/sec³t * (sec²t dt)=∫cost dt= sint + C= tant*cost + C= x/√(x^2+1) + C2)令:x=t^6 ,∫1...
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