求∫1/[x^2根号(x^2+1)]dx的不定积分

2个回答

  • x=tana

    dx=sec²ada

    原式=∫sec²ada/tan²a*seca

    =∫secada/tan²a

    =∫da/sina

    ∫sinada/sin²a

    =∫dcosa/(cos²a-1)

    =(1/2)[∫dcosa/(cosa-1)-∫dcosa/(cosa+1)]

    =(1/2)(ln|cosa-1|-ln|cosa+1|)+C

    =(1/2)ln|(cosa-1)/(cosa+1)|+C

    =(1/2)ln|(cosa-1)²/sin²a|+C

    =ln|(cosa-1)/sina|+C

    =ln|cota-csca|+C

    =ln|1/x-√[1+(1/x)²]|+C

    =ln|[1-√(x²+1)]/x|+C