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x³/x²+2x%3的不定积分

∫x³/(x²+2x-3)dx=∫(x³+2x-3x-2x+3)/(x²+2x-3)dx =∫x+3/(x²+2x-3)dx =∫xdx+3∫1/(x²+2x-3)dx =x²/2+3∫1/[(x-1)(x+1)]dx =x²/2+3/4∫1/(x-1)-1/(x+3)]dx = x²/2+3/4ln|x-1|-3/4ln|x+3|+C

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 ...

求不定积分∫[x³√(4-x²)]dx 解:原式=2∫{x³√[1-(x/2)²]}dx【令x/2=sinu,则x=2sinu,dx=2cosudu,代入原式得】 =32∫sin³ucos²udu=32∫sin³u(1-sin²u)du=32[∫sin³du-∫sin⁵udu] =32{∫sin³d...

令x=3sina dx=3cosada √(9-x²)=3cosa a=arcsin(x/3) sina=x/3 则cosa=√(1-x²/9)=√(9-x²)/3 所以原式=∫(1-3sina)/3cosa*3cosada =∫(1-3sina)da =a+3cosa+C =arcsin(x/3)+√(9-x²)/3+C

原式=∫ (x³-3x²+3x-1)/x² dx =∫ (x-3+3/x-1/x²)dx =x²/2-3x+3ln|x|-1/x+C

多项式除法 =x²+x-4+(6x+8)/(x²+2x+2) 积分=x³/3+x²/2-4x+3∫1/(x²+2x+2)d(x²+2x+2)+∫2/((x+1)²+1)d(x+1) =x³+x²/2-4x+3ln(x²+2x+2)+2arctan(x+1)+C

∫dx/(x²+x+1) =4∫dx/(4x²+4x+1+3) =4∫dx/[(2x+1)²+3] = 4/3∫dx/{[(2x+1)/√3]²+1} = 2/√3∫d[(2x+1)/√3]/{[(2x+1)/√3]²+1} =2arctan[(2x+1)/√3]/√3+C

∫x+5/x²-6x+13 dx =∫(x-3+8)/(x²-6x+13) dx =∫(x-3)/(x²-6x+13)dx+8∫1/[(x-3)²+2²]dx =1/2ln(x²-6x+13)+8/2 arctan(x-3)/2+c =1/2ln(x²-6x+13)+4 arctan(x-3)/2+c

=∫(x+1/2)/(x²+x+1)dx+1/2∫1/((x+1/2)²+3/4)dx =1/2∫1/(x²+x+1)d(x²+x+1)+1/2∫1/(u²+3/4)du =(1/2)ln(x²+x+1)+(1/2)/(3/4)*√3/2*arctan(2u/√3)+C =(1/2)ln(x²+x+1)+(1/√3)arctan((2x+1)/√3)+C

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