求不定积分2x-5 x^2-5x 7
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详见:http://hi.baidu.com/xxllxhdj/blog/item/0f5c8a0c96aab9c762d986c5.html
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∫1/(x^2+2x+5)dx=∫1/[(x+1)^2+4]dx=∫(1/4)/[[(x+1)/2]^2+1]dx=∫(1/4)·2/[[(x+1)/2]^2+1]d((x+1)/2)=(1/2)∫1
令1+x=√6sinu,则:u=arcsin[(1+x)/√6],dx=√6cosudu.∴∫[1/√(5-2x-x^2)]dx=∫{1/√[6-(1+x)^2]}dx=√6∫{1/√[6-6(sin
∫arcsinx/×2DX=-∫arcsinxd(1/x)的=-(1/x)的*arcsinx+∫(1/X)D(arcsinx)=-arcsinx/X+∫(1/X)*[1/√(1-X2)]DXX=圣马丁
原式=∫(x+1)/x²+∫xlnxdx=∫x/x²+∫1/x²+1/2∫lnxdx²=∫1/x+∫1/x²+1/2*x²lnx-1/2∫x
∫2x*sin(x²)dx=∫sin(x²)dx²=-cos(x²)+C
1/(x+1)(x+2)(x+3)=1/(x+1)[1/(x+2)-1/(x+3)]=1/[(x+1)(x+2)]-1/[(x+1)(x+3)]=1/(x+1)-1/(x+2)-1/2[1/(x+1)
∫dx/x^2=∫x^(-2)*dx=1/(-2+1)*x^(-2+1)+C=-1/x+C
先用分部积分法,然后用t=1-5x^2简化,结果记得换回x 我算的结果1/4(1-5x^2)^10*x^4-1/1200x^6+1/1100x^(11/2) 希望对你有用
原式=∫x^(1/2)*(x^2-5)dx=∫[x^(5/2)-5x^(1/2)]dx=2/7*x^(7/2)-10/3*x^(3/2)+C
分部积分法∫xe^x/(1+x)^2dx=-∫xe^xd[1/(1+x)]=-xe^x/(1+x)+∫(1+x)e^x×1/(1+x)dx=-xe^x/(1+x)+∫e^xdx=-xe^x/(1+x)
∫(1/x^2+6x+5)dx=∫(1/[(x+5)(x+1)]dx=1/4∫(1/(x+1)-1/(x+5)dx=1/4[ln(x+1)-ln(x+5)]+C非要用什么换元法的话,令x+1=t,dx
∫dx/(4x-x^2)=∫dx/[x(4-x)]=(1/4)∫[(4-x)+x]/[x/(4-x)]dx=(1/4)∫[1/x+1/(4-x)]dx=(1/4)[ln(x)-ln(4-x)]+C=(
4*x^(1/2)4倍根号X
答:1.原式=∫1/[(x+1)^2+4]dx=1/4∫1/[((x+1)/2)^2+1]dx=1/2*arctan[(x+1)/2]+C2.原式=1/2∫1/x-x^6/(x^7+2)dx=1/2[
解∫x/(x^2)dx=∫1/xdx=ln|x|+C
设(2x²-3x-3)/[(x-1)(x²-2x+5)]=[a/(x-1)]+[(bx+c)/(x²-2x+5)]则2x²-3x-3=a(x²-2x+