作业帮∫1/[x*(x^10+2)]dx∫1/[x*(x^10+2)]dx
来源：学生作业帮助网 编辑：作业帮 时间：2024/07/12 20:41:14
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∫1/[x*(x^10+2)]dx∫1/[x*(x^10+2)]dx
∫1/[x*(x^10+2)]dx
∫1/[x*(x^10+2)]dx
∫1/[x*(x^10+2)]dx∫1/[x*(x^10+2)]dx
令u=x^10,du=10x^9 dx = 10u/x dx,dx = x/(10u) du
∫dx/[x(x^10+2)]
= (1/10)∫du/[u(u+2)] du
= (1/20)∫2/[u(u+2)] du
= (1/20)∫[(u+2)-u]/[u(u+2)] du
= (1/20)∫[1/u - 1/(u+2)] du
= (1/20)[∫du/u - ∫d(u+2)/(u+2)]
= (1/20)[ln|u| - ln|u+2|] + C
= (1/20)ln|x^10| - (1/20)ln|x^10+2| + C
= (1/2)ln|x| - (1/20)ln|x^10+2| + C
∫1/[x*(x^10+2)]dx∫1/[x*(x^10+2)]dx
∫[dx/(e^x(1+e^2x)]dx
∫1+2x/x(1+x)*dx∫1+2x/x(1+x) * dx
∫(x-1)^2dx,
∫x^1/2dx
∫dx/x(1+x)
∫1/x^2+x+1dx
∫1/(x^2+x+1)dx
∫dx/x^2(1-x^2)
∫dx/x^2(1+x^2)
∫X^2/1-x^2 dx.
∫X^2/1-x^2 dx.
∫ (x+arctanx)/(1+x^2) dx
∫（x^2+1/x^4)dx
∫2 -1|x²-x|dx
∫x^2/1+X dx.
∫ xe^x/(1+x)^2 dx
∫dx/[(1-x)x^2]