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∫x%1/x∧2+2x+2Dx

∫ (x+1)/(x-2x+2) dx= ∫ [(1/2)(2x-2)+2]/(x-2x+2) dx= (1/2)∫ (2x-2)/(x-2x+2) dx + ∫ 2/(x-2x+2) dx= (1/2)∫ d(x-2x+2)/(x-2x+2) + 2∫ 1/[(x-1)+1] d(x-1)= (1/2

答案是-x^(-1)+x^2+2x 解释:可以进行分步积分,先求∫1/x^2dx,1/x^2的原函数是-x^(-1),即它的积分为-x^(-1);第二步求解∫2xdx,因为2x的原函数为x^2,即∫2xdx=x^2;第三步求解∫2dx,2的原函数为2x,即∫2dx2x.求积分的大致方法就是对每一个分别求积分希望您能满意~

x^2+2x+2 = (x+1)^2+1, 记 x+1 = tanu, 则 dx = (secu)^2duI = ∫ (secu)^2du/[(tanu)^2+1]^2 = ∫ (secu)^2du/(secu)^4= ∫ (cosu)^2du = (1/2)∫(1+cos2u)du= u/2 + (1/4)sin2u + C = u/2 + (1/2)sinucosu + C = (1/2)arctan(x+1) + (1/2)(x+1)/(x^2+2x+2) + C

∫(x^2-2x)dx/(1+x^2)=∫x^2dx/(1+x^2)-∫2xdx/(1+x^2)=∫1dx(1+x^2)-∫1dx-∫2xdx(1+x^2)=tanx-x-ln(x^2+1)

三角换元来做;有x^2和x^2+1,利用tan换元;过程如下:令x=tanu,则x+1=secu,dx=secudu∫x^2/(x^2+1)^2dx=∫ [tanu/(secu)^4]secudu=∫ tanu/secudu=∫ (secu-1)/secudu=∫ 1 du - ∫ cosu du=u - (1/2)∫ (1+cos2u) du=u - (1/2)u - (1/4)sin2u + C=(1/2)u - (1/2)sinucosu + C=(1/2)arctanx - (1/2)x/(1+x) + C 望采纳!

∫ x/(1 + x) dx,令x = tanz,dx = secz dz= ∫ tanz/secz * (secz dz)= ∫ sinz/cosz * cosz dz= ∫ (1 - cos2z)/2 dz= z/2 - (1/4)sin2z + C= (1/2)arctanx - (1/2) * x/√(1 + x) * 1/√(1 + x) + C= (1/2)arctanx - x/[2(1 + x)] + C

∫(x-2)/(x^2+2x+3)^2 dx= (1/2) ∫ d(x^2+2x+3)/(x^2+2x+3)^2 - ∫3/(x^2+2x+3)^2 dx= -(1/2)(1/(x^2+2x+3)) - ∫3/(x^2+2x+3)^2 dxx^2+2x+3=(x+1)^2+2let x+1 = √2tana dx=√2(seca)^2 da∫1/(x^2+2x+3)^2 dx=∫ (1/[4(seca)^4] )√2(seca)^2 da=(√2/4) ∫ (cosa)

原式=∫(0→1)dx/((x+1)^2+1)=∫(0→1)d(x+1)/((x+1)^2+1)=arctan(x+1)|(0→1)=arctan2-π/4

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

∫ (x^2 + 1)^2 dx= ∫ (x^4 + 2x^2 + 1) dx= (1/5)x^5 + (2/3)x^3 + x + c

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