www.whkt.net > ∫x/1+x^2Dx 求不定积分 求详细过程 谢谢

∫x/1+x^2Dx 求不定积分 求详细过程 谢谢

∫x/(1+x^2)dx=1/2∫1/(1+x^2)d(1+x^2)=1/2ln(1+x^2)+C

lnx+1/3x^3

换元法令x=tany则∫1/[(x^2+1)]^2dx=∫1/secy^4dtany=∫1/secy^2dy=∫cosy^2dy ==∫(cos2y+1)/2dy=y/2-sin2y/4+cy=arctanx所以原式=arctanx/2-sin(2arctanx)/4+c

∫1/(x+1)^3dx=∫(x+1)^(-3)dx=∫(x+1)^(-3)d(x+1)=(-1/2)*(x+1)^(-2)+c.

原式=∫(x+1+2√x)dx=1/2 * x^2 + x + 1/3 * x^(3/2) + C

原式 = ∫((x^4 -1) +1)/(1+x) dx = ∫((x -1)(x+1)/(1+x) +1/(1+x) ) dx =∫ (x-1 + 1/(1+x) ) dx = x/3 -x + arctan x + c

令u=1+x^2则du=2xdx原式=1/2∫1/udu=1/2lnu+c=1/2ln(1+x^2)+c

∫x√1-x^2dx=1/2∫√1-x^2dx^2=-1/2∫√1-x^2d(1-x^2)=-1/3(1-x^2)^(3/2)+C祝开心

x/(1+x)^4 =(x+1-1)/(1+x)^4=1/(1+x)^3 -1/(1+x)^4所以∫x/(1+x)^4 dx=∫1/(x+1)^3 dx -∫1/(x+1)^4 dx=-1/2 (x+1)^(-2) +1/3 (x+1)^(-3) +c

解:∫1/X^2dx=∫(x^-2)dx=x^(-2+1)/-2+1+c=-1/x+c ∫1/x+2dx=ln|1/(x+2)|(x≠-2)+c

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