∫xlndx
∫xlndx 的不定积分是多少 -
∫xlnxdx=(1/2)∫lnxdx^2 =(1/2)x^2lnx-(1/2)∫x^2dlnx =(1/2)x^2lnx-(1/2)∫xdx =(1/2)x^2lnx-(1/4)x^2+C
∫ xlnxdx =1/2*∫lnxdx²=1/2*x²lnx-1/2*∫ x²dlnx =1/2*x²lnx-1/2*∫ x²*dx/x =1/2*x²lnx-1/4*x²=(2lnx-1)x²/4
∫xlnxdx=(1/2)∫lnxdx^2 =(1/2)x^2lnx-(1/2)∫x^2dlnx =(1/2)x^2lnx-(1/2)∫xdx =(1/2)x^2lnx-(1/4)x^2+C
∫ xlnxdx =1/2*∫lnxdx²=1/2*x²lnx-1/2*∫ x²dlnx =1/2*x²lnx-1/2*∫ x²*dx/x =1/2*x²lnx-1/4*x²=(2lnx-1)x²/4
= 0.5x^2 *lnx - ∫ 0.5x dx = 0.5x^2 *lnx - 0.25x^2 +C，C为常数，
∫xln(1+x^2)dx =1/2∫ln(1+x^2)dx^2 =1/2∫ln(1+x^2)d(1+x^2)=1/2(1+x^2)ln(1+x^2)-1/2∫(1+x^2)dln(1+x^2)=1/2(1+x^2)ln(1+x^2)-1/2∫(1+x^2)*1/(1+x^2)d(1+x^2)=1/2(1+x^2)ln(1+x^2)-1/2∫dx^2 =1/2(1+x^2)ln(1+x到此结束了？。
∫xln( x-1) dx的积分过程是什么? -
解答过程如下：利用分部积分法可求得∫xln（x-1）dx =1/2x²ln(1+x)-1/2[x²/2-x+ln(1+x)]+C∫x ln(x-1)dx=x^2/2* ln(x-1)-∫x^2/2ln(x-1)'dx =x^2/2* ln(x-1)-∫x^2/2(x-1)dx =x^2/2* ln(x-1)-∫(x^2-x)/2(x-1)dx-∫x/2(x-1到此结束了？。
∫xln（x-1）dx=1/2x²ln(1+x)-1/2[x²/2-x+ln(1+x)]+C。C为积分常数。解答过程如下：∫xln（x-1）dx =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)等会说。
∫xln(1+ x) d=什么? -
∫xln(1+x)dx,令u=x+1=∫(u-1)*lnu du=∫ulnu du-∫lnu du=∫lnu d(u²/2)-(ulnu-∫ du)=u²/2*lnu-∫u²/2 d(lnu)-ulnu+u=u²/2*lnu-1/2*∫u du-ulnu+u=u²/2*lnu-1/4*u²-ulnu+u+C=(1/2)(x+1)²ln(x+1)-还有呢？
简单分析一下，详情如图所示，
∫xln(1+x)dx -
解答过程如下：∫xln（x-1）dx =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+后面会介绍。
解答如图：