Question

关于不定积分的一道题∫（1/((x+1)^2*(√(x^2+2x+2)))dx.

Answer 1

令 x+1=tanu,则· dx=(secu)^2du,

I=∫dx/[(x+1)^2*√(x^2+2x+1)] =∫dx/{(x+1)^2*√[(x+1)^2+1]}

=∫(secu)^2du/[(tanu)^2*secu] =∫secudu/(tanu)^2

=∫cosudu/(sinu)^2=∫dsinu/(sinu)^2=C-1/sinu=C-cscu

=C-√(x^2+2x+2)/(x+1).