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∫√x^2+9/x Dx 的不定积分

令t=√(x^2-9), t^2=x^2-9, 2tdt=2xdx tdt=xdx 积分号下:√(x^2-9)dx/x =√(x^2-9) xdx/x^2 (分子分母同乘以x) =t *tdt/(t^2+9) =t^2dt/(t^2+9) =[1-9/(t^2+9)]dt ∫[1-9/(t^2+9)]dt=t-3arctan(t/3)+C=√(x^2-9)-3arctan[√(x^2-9)/3]+C 希望对你能...

令x=3tant 则原式=3∫csctdtant =3csct·tant+3∫tant·csctcottdt =3sect+3∫cscdt =3sect+3ln|csct-cott|+C =√(x^2+9)+3ln|√(x^2+9)-3|-3ln|x|+C

令√(x^2-9)=u,则:x^2=u^2+9,∴d(x^2)=2udu。 ∴∫[√(x^2-9)/x]dx =(1/2)∫[2x√(x^2-9)/x^2]dx =(1/2)∫[√(x^2-9)/x^2]d(x^2) =(1/2)∫[u/(u^2+9)]·2udu =∫{[(u^2+9)-9]/(u^2+9)}du =∫du-9∫...

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∫ √(9 - x²) dx --> x = 3 siny,dx = 3 cosy dy --> = ∫ √(9 - 9sin²y) 3cosy dy = ∫ 3cosy 3cosy dy = 9∫ cos²y dy = (9/2)∫ (1 + cos2y) dy = (9/2)[y + (1/2)sin2y] + C = (9/2)y + (9/2)sinycosy + C = (9/2)arcsin(x/3) + ...

您好,答案如图所示:

∫√(x²-9)/xdx=√(x²-9)-3arc sec(x/3)+C

记t=√(x+9),则x=t²-9 不定积分可化为∫t/(t²-9)d(t²-9)dt=∫2t²/(t²-9)dt=∫[2+3/(t-3)-3/(t+3)]dt=2t+3ln(t-3)-3ln(t+3)

令t=9-x² => dt=-2xdx ∫x³/√(9-x²) dx = ∫(9-t)x/√t * -1/(2x) dt = (-1/2)∫(9-t)/√t dt = (-1/2)∫(9/√t - √t) dt = (1/2)∫√t dt - (9/2)∫1/√t dt = (1/3)t^(3/2) - 9√t + C = (1/3)(9-x²)^(3/2) - 9√(9-x²) + C = (-1...

我想你的题应该是这样吧 ∫ x³/(9+x²) dx =(1/2)∫ x²/(9+x²) d(x²) =(1/2)∫ (x²+9-9)/(9+x²) d(x²) =(1/2)∫ 1 d(x²) - (9/2)∫ 1/(9+x²) d(x²) =(1/2)x² - (9/2)ln(x²+9) + C ...

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