Question

高数 不定积分 有理函数的积分

Answer 1

1、原式=(1/2)*∫d(x^2-2x+5)/(x^2-2x+5)+2∫dx/(x^2-2x+5)
=(1/2)*ln|x^2-2x+5|+2∫dx/[(x-1)^2+4]
=(1/2)*ln|x^2-2x+5|+arctan[(x-1)/2]+C，其中C是任意常数
2、令u=tan(x/2)，则cosx=(1-u^2)/(1+u^2)，dx=2du/(1+u^2)
原式=∫1/[3+(1-u^2)/(1+u^2)]*2du/(1+u^2)
=∫du/(2-u^2)
=(√2/4)*∫[1/(√2-u)+1/(√2+u)]du
=(√2/4)*[ln|√2+u|-ln|√2-u|]+C
=(√2/4)*[ln|√2+tan(x/2)|-ln|√2-tan(x/2)|]+C，其中C是任意常数

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1、原式=(1/2)*∫d(x^2-2x+5)/(x^2-2x+5)+2∫dx/(x^2-2x+5)
=(1/2)*ln|x^2-2x+5|+2∫dx/[(x-1)^2+4]
=(1/2)*ln|x^2-2x+5|+16arctan[(x-1)/2]+C