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∫x%1/x∧2+2x+2Dx

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+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)ln| x-2x+2 | + 2arctan(x-1) + CNote:Let x+1 = A(2x-2)+Bx+1 = 2Ax + (B-2A)2A=1 => A=1/2B-2A=1 => B=1+2(1/2)=2∴x+1 = (1/2)(2x-2)+2

^x^2+2x+2 = (x+1)^自2+1, 记知 x+1 = tanu, 则 dx = (secu)^2du I = ∫道 (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

∫(0,-2)1/x∧2+2x+2dx=∫(0,-2)1/[(x+1)^2+1]dx=arctan(x+1)|(0,-2)=arctan(-2+1)-arctan(0+1)=-π/4-π/4=-π/2

你的表达可能有点问题,是不是想求 ∫[(x+1)/(x^2+x+2)]dx ?若是这样,则方法如下:原式=∫{[(x+1/2)+1/2]/[(x+1/2)^2+7/4]}dx =∫{(x+1/2)/[(x+1/2)^2+7/4]}d(x+1/2) +(1/2)∫{1/[(x+1/2)^2+7/4]}d(x+1/2)令x+1/2=u,则:x=u-1/2,dx=du.∴原式=∫[u/(u^2+7/4

你好∫负无穷到正无穷1/(x^2+2x+2)dx=∫负无穷到正无穷1/[(x+1)^2+1]dx=∫负无穷到正无穷1/1/[(x+1)^2+1]d(x+1)=arctan(x+1)│负无穷到正无穷=π/2-(-π/2)=π【数学辅导团】为您解答,不理解请追问,理解请及时选为满意回答!(*^__^*)谢谢!

∫(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)

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-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)

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!

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