求定积分∫xlnxdx

璩采浅732求定积分∫(上限为e平方,下限为e)1/x乘以(lnx)平方dx -
茹省馥19569415208 ______ 根据题意,先求不定积分部分:∫(lnx)^2/x dx=∫(lnx)^2 d(lnx)=(1/3)(lnx)^3.所以,则定积分为:定积分=(1/3){[ln(e^2)]^3-[lne]^3}=(1/3)(8-1)=7/3.

璩采浅732X乘以lnX.的积分 -
茹省馥19569415208 ______[答案] 分部积分即可: ∫xlnxdx =(1/2)∫lnxd(x^2) =(1/2)x^2lnx-(1/2)∫x^2dlnx =(1/2)x^2lnx-(1/2)∫x^2·(1/x)dx =(1/2)x^2lnx-(1/2)∫xdx =(1/2)x^2lnx-(1/4)x^2

璩采浅732求下列不定积分或定积分 1,∫(x^4+cosx)dx 2,∫cosxsin^3xdx 3,∫xlnxdx 4,∫|x - 3|dx求下列不定积分或定积分 1,∫(x^4+cosx)dx 2,∫cosxsin^3xdx 3,∫xlnxdx 4,∫|x - 3|dx -
茹省馥19569415208 ______[答案] 1、 ∫(x^4+cosx) dx = (1/5)x^5+sinx + C 2、 ∫(cosxsin³x) dx = ∫sin³x d(sinx) = (1/4)sin^4x + C 3、 ∫xlnx dx = ∫lnx d(x²/2) = (1/2)x²lnx - (1/2)∫x²*1/x dx = (1/2)x²lnx - (1/2)∫x dx = (1/2)x²lnx - (1/4)x² + C 4、 |x-3|=[√(x-3)]² =-(x-3)=-x+3 for x= x-3 for ...

璩采浅732xlnx的积分怎么求 -
茹省馥19569415208 ______ ∫xlnxdx=(1/2)x²lnx-(1/4)x²+C.(C为积分常数) 解答过程如下: ∫xlnxdx =(1/2)∫lnxd(x²) =(1/2)x²lnx-(1/2)∫x²*(1/x)dx =(1/2)x²lnx-(1/2)∫xdx =(1/2)x²lnx-(1/4)x²+C 扩展资料: 分部积分: (uv)'=u'v+uv',得:u'v=(uv)'-uv'. 两边积分得:∫ u'...

璩采浅732计算定积分∫e(在上)1(在下)xlnxdx(在中间) -
茹省馥19569415208 ______[答案] 用凑微分和分部积分的方法做此题具体步骤如下:∫(上限e)(下限1)xlnxdx=∫(上限e)(下限1)lnxd((x^2)/2)=1/2*x^2*lnx|(上限e)(下限1)-∫(上限e)(下限1)((x^2)/2)d(lnx)=1/2*e^2-∫(上限e)(下限1)1/2*xdx=1/2*e^2-1/4*...

璩采浅732求定积分∫( - 1~0) √(1 - x2) dx= -
茹省馥19569415208 ______ 用定积分几何意义求 被积函数为y=√(1-x²), 化成圆的方程 y²=1-x² 即x²+y²=(1)²所以 此定积分表示的曲线是圆心在原点,半径为1的1/4圆周.所以定积分为π*1²/4=π/4

璩采浅732计算下列定积分:∫上限1下限0(xe^x)dx; ∫上限1e下限0xlnxdx;求过程! -
茹省馥19569415208 ______[答案] ∫(0→1) xe^x dx = ∫(0→1) x d(e^x) = xe^x - ∫(0→1) e^x dx = [(1)e^(1) - (0)e^(0)] - e^x = e - [e^(1) - e^(0)] = e - e + 1 = 1 ∫(0→e) xlnx dx = ∫(0→e) lnx d(x²/2) = (1/2)x²lnx - (1/2)∫(0→e) x² d(lnx) = [(1/2)(e²)ln(e) - (1/2)(0)] - (1/2)∫(0→e) x dx = (1/2)e² - (1/2...

璩采浅732∫(0→4)arctan√xdx
茹省馥19569415208 ______ 用分部积分就可以求了. ∫(0→4)arctan√xdx =xarctan√x(0→4)-∫(0→4)x/(1-x)dx =4arctan2+∫(0→4)x/(1+x)dx =4arctan2+∫(0→4)[1-(1/(1+x))]dx =4arctan2+x(0→4)-ln(1+x)(0→4) =4arctan2+2-ln5

璩采浅732求定积分上限e下限1,xln xdx,上限e - 1下限1,ln(1+x)dx -
茹省馥19569415208 ______[答案] 定积分上限e下限1,xln xdx, =∫(1,e)lnxd(x^2)/2 x^2/2*lnx|(1,e)-∫(1,e)(x^2)/2dlnx =e^2/2-x^2/4|(1,e) =e^2/2-e^2/4+1/4 =1/4*(e^2+1) 上限e-1下限1,ln(1+x)dx =xln(1+x)|(1,e-1)-∫(1,e-1)xdln(1+x) =(e-1)-ln2-∫(1,e-1)x/(x+1)dx =(e-1)-ln2-∫(1,e-1)(x+1-1)/(x+1)dx =e-1...

璩采浅732求定积分∫(1,0)xln(x+1)dx -
茹省馥19569415208 ______[答案] 求定积分[0,1]∫xln(x+1)dx原式=[0,1](1/2)∫ln(x+1)dx²=[0,1](1/2){x²ln(x+1)-∫[x²/(x+1)]dx}=[0,1](1/2){x²ln(x+1)-∫[(x-1)+1/(x+1)]dx}=[0,1](1/2){x²ln(x+1)-(x-1)²/2-ln(x+1)}...