计算不定积分∫xarctanxdx
∫xarctanxdx=x²\/2arctanx-1\/2x+1\/2arctanx+c。c为积分常数。解答过程如下:∫xarctanxdx =∫arctanxdx²\/2 =x²\/2arctanx-∫x²\/2darctanx =x²\/2arctanx-1\/2∫x²\/(1+x²)dx =x²\/2arctanx-1\/2∫(x²+1-1)\/(1+x...
xarctanx不定积分怎么算?
xarctanx不定积分:∫xarctanxdx=∫arctanxd(x²\/2)=(x²\/2)arctanx-(1\/2)∫x²d(arctanx)=(1\/2)x²arctanx-(1\/2)∫x²\/(x²+1)dx=(1\/2)x²arctanx-(1\/2)∫[(x²+1)-1]\/(x²+1)dx=(1\/2)x²arctanx-(...
xarctanxdx的不定积分是什么?
xarctanx不定积分:∫xarctanxdx =∫arctanxd(x²\/2)=(x²\/2)arctanx-(1\/2)∫x²d(arctanx)=(1\/2)x²arctanx-(1\/2)∫x²\/(x²+1)dx =(1\/2)x²arctanx-(1\/2)∫[(x²+1)-1]\/(x²+1)dx =(1\/2)x²arctan...
xarctanxdx的定积分0到1是什么?
∫xarctanxdx=1\/2 ∫arctanxdx^2 =1\/2[x^2arctanx|(0,1)-∫(0,1)x^2\/(1+x^2)dx]=1\/2[π\/4-∫(0,1)1-1\/(1+x^2)dx]=1\/2[π\/4-∫(0,1)dx+∫(0,1)1\/(1+x^2)dx]=1\/2[π\/4-x|(0,1)+arctanx|(0,1)]=π\/4-1\/2 积分基本公式 1、∫0dx=c 2、...
计算不定积分=∫xarctanxdx
∫xarctanxdx =1\/2∫arctanx*2xdx =1\/2∫arctanxdx^2 =1\/2xarctanx-1\/2∫x^2*1\/(x^2+1)dx =1\/2xarctanx-1\/2∫(x^2+1-1)dx\/(x^2+1)=1\/2xarctanx-1\/2∫dx+1\/2∫dx\/(x^2+1)=1\/2xarctanx-x\/2+1\/2*arctanx+C =1\/2*(xarctanx-x+arctanx)+C 黎曼积分...
求不定积分∫xarctanxdx
不定积分的结果,会随着方法不同而结果不同,其实都是对的
积分问题∫xArctanx dx=?? & ∫xArcsinx dx =?
∫xArctanx dx =1\/2积分:arctanxdx^2 =x^2\/2arctanx-1\/2积分:x^2d(arctanx)=x^2\/2arctanx-1\/2积分:x^2\/(1+x^2)dx =x^2\/2arctanx-1\/2积分:(x^2+1-1)\/(1+x^2)dx =x^2\/2arctanx-x\/2+arctanx+C ∫xArcsinx dx 同样的方法:=1\/2积分:arcsinxdx^2 =x^2\/2...
∫xarctanxdx用换元法怎么做 ??
= ∫ u tanu d(tanu)= (1\/2)∫ u d(tan²u)= (1\/2)u tan²u - (1\/2)∫ tan²u du = (1\/2)u tan²u - (1\/2)∫ (sec²u - 1) du = (1\/2)u tan²u - (1\/2)(tanu - u) + C = (1\/2)x² arctanx - (1\/2)x ...
计算∫xarctanxdx属于第二类换元法求解吗
不知道算不算,但可以使用分部积分法
用部分积分法求下列不定积分:∫xarctan xdx 要过程。。
∫xarctanxdx =∫arctanxd(0.5*x^2)=0.5*x^2 *arctanx-∫0.5*x^2d(arctanx)=0.5*x^2 *arctanx-∫0.5*x^2\/(1+x^2)dx =0.5*x^2 *arctanx-0.5*∫(1-(1\/(1+x^2))dx =0.5*x^2 *arctanx-0.5*∫dx+0.5*∫(1\/(1+x^2))dx =0.5*x^2 *arctanx...