1\/(x^2+1)(x^2+x)的不定积分怎么算,求大神
∵(x²+1)(x²+x)=(x²+1)x(x+1)，∴设1\/[(x²+1)(x²+x)]=a\/x+b\/(x+1)+(cx+d)\/(x²+1)。解得a=1，b=-1\/2，c=-1\/2，d=-1\/2。∴原式=ln丨x丨-(1\/2)ln丨x+1丨-(1\/4)ln(x²+1)-(1\/2)arctanx+C。供参考。
求∫dx\/((x^2+1)(x^2+x)不定积分
1\/(x^2+1)(x^2+x)=[(ax+b)\/(x^2+1)]+(c\/x)+[d\/(x+1)]右边通分对应项相等，即可得到：a=b=d=-1\/2,c=1.此时积分为：原式 =-（1\/2)∫(x+1)dx\/(x^2+1)+∫dx\/x-(1\/2)∫dx\/(x+1)=-(1\/2)∫xdx\/(x^2+1)-(1\/2)∫dx\/(1+x^2)-lnx-(1\/2)ln(x+1)...
求∫dx\/((x^2+1)(x^2+x)不定积分
1\/(x^2+1)(x^2+x)=[(ax+b)\/(x^2+1)]+(c\/x)+[d\/(x+1)]右边通分对应项相等，即可得到：a=b=d=-1\/2,c=1.此时积分为：原式 =-（1\/2)∫(x+1)dx\/(x^2+1)+∫dx\/x-(1\/2)∫dx\/(x+1)=-(1\/2)∫xdx\/(x^2+1)-(1\/2)∫dx\/(1+x^2)-lnx-(1\/2)ln(x+1)...
dx\/[(x^2+1)(x^2+x)]不定积分?
朋友，您好！此题非常简单，主要就是待定系数法做，详细过程如图rt所示，希望能帮到你解决问题
1\/[(x^2+1)(x^2+x)]求不定积分
点击放大，荧屏放大再放大：
∫dx\/((x^2+1)(x^2+x)
∫dx\/((x^2+1)(x^2+x)的不定积分是ln│x│-(1\/2)ln│x+1│-(1\/4)ln(x²+1)-(1\/2)arctanx+C。解：∫dx\/((x^2+1)(x^2+x)dx =∫[1\/x-(1\/2)\/(x+1)-(x\/2)\/(x²+1)-(1\/2)\/(x²+1)]dx =ln│x│-(1\/2)ln│x+1│-(1\/4)ln(x&#...
求1\/(x^2+1)(x^2+x+1)的不定积分,谢谢
先拆成两项 然后用积分表里的公式，详情如图所示 关于公式的用法如图
求不定积分,lnx除以[(1+x^2)]的2分之3次dx ,1\/[(x^2+1)(x^2+x)]dx
= xlnx\/√(1 + x²) - ln|x + √(1 + x²)| + C ∫ 1\/[(x² + 1)(x² + x)] dx = ∫ 1\/[x(x + 1)(x² + 1)] dx = ∫ 1\/x dx - (1\/2)∫ dx\/(x + 1) - (1\/2)∫ x\/(x² + 1) dx - (1\/2)∫ dx\/(x...
求∫1\/[x(x^2+1)]dx的不定积分
法一：令u=x^2,则du=2xdx ∫1\/[x(x^2+1)]dx =1\/2·∫1\/[u(u+1)]du =1\/2·∫[1\/u-1\/(u+1)]du =1\/2·∫1\/u du-1\/2·∫1\/(u+1) d(u+1)=1\/2·lnu-1\/2·ln(u+1)+C =1\/2·ln[u\/(u+1)]+C =1\/2·ln[x^2\/(x^2+1)]+C 法二：∫1\/[x(x^2...
1\/(x^2+1)^2的不定积分怎么算
∫（1\/（x^2+1）^2）dx的不定积分为1\/2*x\/(1+x^2)+1\/2arctanx+C。解：令x=tant，则t=arctanx，且x^2+1=(tant)^2+1=(sect)^2 ∫（1\/（x^2+1）^2）dx =∫（1\/(sect)^4）dtant =∫（(sect)^2\/(sect)^4）dt =∫（1\/(sect)^2）dt =∫(cost)^2dt =1\/2∫(...
