1、∫[2/(1+x^2) -sec^2)dx2、∫[√(x)+1]^2/(x) dx

网友回答

∫[2/(1+x^2) -(secx)^2]dx
=∫2/(1+x^2)dx - ∫(secx)^2dx
=2∫1/(1+x^2)dx - tanx + c
=2arctanx - tanx + c
∫[√(x)+1]^2/(x) dx
令√x=t,x=t^2,dx=2tdt
原式=∫(t+1)^2 / t^2 * 2tdt
=2∫(t + 2 + 1/t)dt
=2(1/2 * t^2 + 2t + lnt)
=x + 4√x + 2ln√x + c
======以下答案可供参考======
供参考答案1:
2∫[√x+1)^2dx/x
=∫(1+2/√x+1/x)dx
=x+4√x+lnx+C
1∫[2/(1+x^2)-secx^2]dx
=2arctanx-tanx+C