∫ln+x+1+dx
@濮庭5631：求不定积分? ∫ ln(x+1) dx -
盛到19257127526…… ∫ln(x+1)dx=∫ln(x+1)d(x+1)=(ln(x+1))(x+1)-∫(x+1) d(ln(x+1)) =(x+1)ln(x+1)-∫((x+1)/(x+1))dx=(x+1)ln(x+1)-x+c
@濮庭5631：∫ln(x*x+1)dx -
盛到19257127526…… ∫ln(x*x+1)dx=xln(x^2+1)-Sxdln(x^2+1)dx=xln(x^2+1)-S2x^2/(x^2+1)dx =xln(x^2+1)-2S(1-1/(x^2+1))dx=xln(x^2+1)-2x+2arctanx+c
@濮庭5631：谁知道不定积分∫xln(x+1)dx是多少啊? -
盛到19257127526…… ∫xln(x-1)dx 利用分部积分法: =1/2∫ln(1+x)dx² =1/2x²ln(1+x)-1/2∫x²dln(1+x) =1/2x²ln(1+x)-1/2∫x²/(1+x) dx 分解多项式,变换积分形式: =1/2x²ln(1+x)-1/2∫(x²-1+1)/(1+x) dx =1/2x²ln(1+x)-1/2∫[(x²-1)/(x+1)+1/(1+x)] dx =1/2x²ln(1+x)-1/2∫[(x...
@濮庭5631：∫1/(x² +x +1)dx怎么算 -
盛到19257127526…… ∫√(1+x²) dx=√(1+x²) *x-∫x*d√(1+x²) =√(1+x²) *x-∫x*x/√(1+x²)dx=√(1+x²) *x-∫(x²+1-1)/√(1+x²)dx=√(1+x²) *x-∫[√(x²+1)-1/√(1+x²)]dx=√(1+x²) *x-∫√(x²+1)dx+∫1/√(1+x²)dx 移相 所以2*∫√(1+x²) dx=√(1+x²) *x+∫1/√(1+x²...
@濮庭5631：求不定积分∫xln(x+1)dx -
盛到19257127526…… ∫xln(x+1)dx =∫ln(x+1)d(1/2*x^2) =1/2*x^2*ln(x+1)-1/2*∫x^2dln(x+1) =1/2*x^2*ln(x+1)-1/2*∫x^2/(x+1)dx =1/2*x^2*ln(x+1)-1/2*∫[x-1+1/(x+1)]dx =1/2*x^2*ln(x+1)-1/2*[1/2*x^2-x+ln(x+1)]+C =1/2*(x^2-1)*ln(x+1)-1/4*(x^2-2x)+C
@濮庭5631：∫ln(x+1)dx -
盛到19257127526…… ∫ ln(x + 1) dx= xln(x + 1) - ∫ x d[ln(x + 1)]= xln(x + 1) - ∫ x/(x + 1) dx= xln(x + 1) - ∫ (x + 1 - 1)/(x + 1) dx= xln(x + 1) - ∫ [1 - 1/(x + 1)] dx= xln(x + 1) - x + ln|x + 1| + C
@濮庭5631：∫xln(1+x)dx - 作业帮
盛到19257127526…… [答案] 原式=1/2∫ln(1+x)dx² =1/2x²ln(1+x)-1/2∫x²dln(1+x) =1/2x²ln(1+x)-1/2∫x²/(1+x) dx =1/2x²ln(1+x)-1/2∫(x²-1+1)/(1+x) dx =1/2x²ln(1+x)-1/2∫[(x²-1)/(x+1)+1/(1+x)] dx =1/2x²ln(1+x)-1/2∫[(x-1)+1/(1+x)] dx =1/2x²ln(1+x)-1/2[x²/2-x+ln(1+x)]+C
@濮庭5631：求∫xln(x+1)dx. - 作业帮
盛到19257127526…… [答案] 据题,有 ∫xln(x+1)dx = x2 2ln(x+1)−∫ x2 2• 1 x+1dx = x2 2ln(x+1)− 1 2∫(x−1+ 1 x+1)dx = x2 2ln(x+1)− 1 2( x2 2−x+ln(x+1))+C.
@濮庭5631：求不定积分∫xln(x+1)dx - 作业帮
盛到19257127526…… [答案] ∫xln(x+1)dx=∫ln(x+1)d(1/2*x^2)=1/2*x^2*ln(x+1)-1/2*∫x^2dln(x+1)=1/2*x^2*ln(x+1)-1/2*∫x^2/(x+1)dx=1/2*x^2*ln(x+1)-1/2*∫[x-1+1/(x+1)]dx=1/2*x^2*ln(x+1)-1/2*[1/2*x^2-x+ln(x+1...
@濮庭5631：∫ln(x+1)dx怎么解 -
盛到19257127526…… 分步积分u=x,v=ln(x+1),u导=1,v导=1/(x+1)ln(x+1)dx=xln(x+1)-∫[x/(x+1)]dx=xln(x+1)-∫[1-1/(x+1)]dx=xln(x+1)-∫dx+∫1/(x+1)dx=xln(x+1)-x+ln(x+
盛到19257127526…… ∫ln(x+1)dx=∫ln(x+1)d(x+1)=(ln(x+1))(x+1)-∫(x+1) d(ln(x+1)) =(x+1)ln(x+1)-∫((x+1)/(x+1))dx=(x+1)ln(x+1)-x+c
@濮庭5631：∫ln(x*x+1)dx -
盛到19257127526…… ∫ln(x*x+1)dx=xln(x^2+1)-Sxdln(x^2+1)dx=xln(x^2+1)-S2x^2/(x^2+1)dx =xln(x^2+1)-2S(1-1/(x^2+1))dx=xln(x^2+1)-2x+2arctanx+c
@濮庭5631：谁知道不定积分∫xln(x+1)dx是多少啊? -
盛到19257127526…… ∫xln(x-1)dx 利用分部积分法: =1/2∫ln(1+x)dx² =1/2x²ln(1+x)-1/2∫x²dln(1+x) =1/2x²ln(1+x)-1/2∫x²/(1+x) dx 分解多项式,变换积分形式: =1/2x²ln(1+x)-1/2∫(x²-1+1)/(1+x) dx =1/2x²ln(1+x)-1/2∫[(x²-1)/(x+1)+1/(1+x)] dx =1/2x²ln(1+x)-1/2∫[(x...
@濮庭5631：∫1/(x² +x +1)dx怎么算 -
盛到19257127526…… ∫√(1+x²) dx=√(1+x²) *x-∫x*d√(1+x²) =√(1+x²) *x-∫x*x/√(1+x²)dx=√(1+x²) *x-∫(x²+1-1)/√(1+x²)dx=√(1+x²) *x-∫[√(x²+1)-1/√(1+x²)]dx=√(1+x²) *x-∫√(x²+1)dx+∫1/√(1+x²)dx 移相 所以2*∫√(1+x²) dx=√(1+x²) *x+∫1/√(1+x²...
@濮庭5631：求不定积分∫xln(x+1)dx -
盛到19257127526…… ∫xln(x+1)dx =∫ln(x+1)d(1/2*x^2) =1/2*x^2*ln(x+1)-1/2*∫x^2dln(x+1) =1/2*x^2*ln(x+1)-1/2*∫x^2/(x+1)dx =1/2*x^2*ln(x+1)-1/2*∫[x-1+1/(x+1)]dx =1/2*x^2*ln(x+1)-1/2*[1/2*x^2-x+ln(x+1)]+C =1/2*(x^2-1)*ln(x+1)-1/4*(x^2-2x)+C
@濮庭5631：∫ln(x+1)dx -
盛到19257127526…… ∫ ln(x + 1) dx= xln(x + 1) - ∫ x d[ln(x + 1)]= xln(x + 1) - ∫ x/(x + 1) dx= xln(x + 1) - ∫ (x + 1 - 1)/(x + 1) dx= xln(x + 1) - ∫ [1 - 1/(x + 1)] dx= xln(x + 1) - x + ln|x + 1| + C
@濮庭5631：∫xln(1+x)dx - 作业帮
盛到19257127526…… [答案] 原式=1/2∫ln(1+x)dx² =1/2x²ln(1+x)-1/2∫x²dln(1+x) =1/2x²ln(1+x)-1/2∫x²/(1+x) dx =1/2x²ln(1+x)-1/2∫(x²-1+1)/(1+x) dx =1/2x²ln(1+x)-1/2∫[(x²-1)/(x+1)+1/(1+x)] dx =1/2x²ln(1+x)-1/2∫[(x-1)+1/(1+x)] dx =1/2x²ln(1+x)-1/2[x²/2-x+ln(1+x)]+C
@濮庭5631：求∫xln(x+1)dx. - 作业帮
盛到19257127526…… [答案] 据题,有 ∫xln(x+1)dx = x2 2ln(x+1)−∫ x2 2• 1 x+1dx = x2 2ln(x+1)− 1 2∫(x−1+ 1 x+1)dx = x2 2ln(x+1)− 1 2( x2 2−x+ln(x+1))+C.
@濮庭5631：求不定积分∫xln(x+1)dx - 作业帮
盛到19257127526…… [答案] ∫xln(x+1)dx=∫ln(x+1)d(1/2*x^2)=1/2*x^2*ln(x+1)-1/2*∫x^2dln(x+1)=1/2*x^2*ln(x+1)-1/2*∫x^2/(x+1)dx=1/2*x^2*ln(x+1)-1/2*∫[x-1+1/(x+1)]dx=1/2*x^2*ln(x+1)-1/2*[1/2*x^2-x+ln(x+1...
@濮庭5631：∫ln(x+1)dx怎么解 -
盛到19257127526…… 分步积分u=x,v=ln(x+1),u导=1,v导=1/(x+1)ln(x+1)dx=xln(x+1)-∫[x/(x+1)]dx=xln(x+1)-∫[1-1/(x+1)]dx=xln(x+1)-∫dx+∫1/(x+1)dx=xln(x+1)-x+ln(x+
