∫arctanx\/(x^2)dx求不定积分
∫arctanx\/(x^2)dx =-∫arctanxd(1\/x)=-arctanx\/x+∫1\/xdarctanx =-arctanx\/x+∫1\/x*1\/(1+x^2)dx =-arctanx\/x+∫[1\/x-x\/(1+x^2)]dx =-arctanx\/x+lnx-1\/2ln(1+x^2)+C
∫(0到+无穷)arctanx\/(x^2)dx= 详细过程 谢谢
dx\/(x^2)=d(-x^(-1))先看不定积分 ∫arctanx\/(x^2)dx =∫arctanxd(-x^(-1))=[-x^(-1)]arctanx-∫[-x^(-1)]d(arctanx)=-arctanx\/x+∫dx\/[x(x^2+1)]=-arctanx\/x+∫[1\/x-x\/(x^2+1)]dx =-arctanx\/x+lnx-(1\/2)ln(x^2+1)在x=0处,得到-无穷 所...
∫(0到+无穷)arctanx\/(x^2)dx=
dx\/(x^2)=d(-x^(-1))先看不定积分 ∫arctanx\/(x^2)dx =∫arctanxd(-x^(-1))=[-x^(-1)]arctanx-∫[-x^(-1)]d(arctanx)=-arctanx\/x+∫dx\/[x(x^2+1)]=-arctanx\/x+∫[1\/x-x\/(x^2+1)]dx =-arctanx\/x+lnx-(1\/2)ln(x^2+1)在x=0处,得到-无穷 所...
反正切x除以x平方的不定积分
∫arctg(x)\/x^2dx =-∫arctg(x)d(1\/x)=-arctg(x)(1\/x)+∫1\/xdarctg(x)=-arctg(x)\/x+∫1\/(x (1+x^2))dx =-arctg(x)\/x+∫1\/x -x\/(1+x^2))dx =-arctg(x)\/x+lnx-ln(1+x^2)\/2+C
不定积分∫arc tanx ÷ x dx 求解
你好!该积分应为ln|x|arctanx-ln|x|+1\/2ln(x²+1)+C其中C为常数,你懂的吧!望采纳!
求arctanx\/(x^2(1+x^2))dx的不定积分
简单计算一下即可,答案如图所示
arctanx\/x²的不定积分怎么求???
结果为:-arctanx\/x+ln丨x丨-(1\/2)ln(1+x²)+C 解题过程如下:解:原式=∫arctanxdx\/x²=∫arctanxd(-1\/x)=-arctanx\/x+∫dx\/[x(1+x²)]=∫dx\/[x(1+x²)]=∫[1\/x-x\/(1+x²)]dx =ln丨x丨-(1\/2)ln(1+x²)+C ∴∫arctanxdx\/...
求不定积分∫x^2arctanxdx
+x-x)\/(1+x²)dx =x³\/3artanx-1\/3∫[x-x\/(1+x²)]dx =x³\/3artanx-x²\/6+1\/6∫1\/(1+x²)d(1+x²)=x³\/3arctanx-x²\/6+1\/6ln(1+x²)+c 解法分析:利用分部积分法求解,进行配凑就可以很快得出结果。
高数题1\/x^2arctanx的不定积分
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