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求不定积分∫xDx/√(x-3)和∫√(x^2-9)Dx/x

∫√(x²-9)/xdx=√(x²-9)-3arc sec(x/3)+C

无法正常回答

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∫...

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...

用凑微分法如图计算,答案与x的符号有关。请采纳,谢谢!

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 ...

您好,答案如图所示:

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=3sint x:-3→3,则t:-π/2→π/2 ∫[-3:3][x+√(9-x²)]dx =∫[-π/2:π/2][3sint+√(9-9sin²t)]d(3sint) =2∫[0:π/2]9cos²tdt =9∫[0:π/2](1+cos2t)dt =9(t+½sin2t)|[0:π/2] =9[(π/2 +½sinπ)-(0+½sin0)] =9π/2

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