Question

求解：∫dx／(1+x²)求不定积分

Answer 1

令x = tany，dx = sec²y dy
∫ dx/(1 + x²)
= ∫ (sec²y dy)/(1 + tan²y)
= ∫ (sec²y dy)/(sec²y)，恒等式：1 + tan²y = sec²y
= ∫ dy
= y + C
= arctan(x) + C

Answer 2

∫dx／(1+x²)可令x=tant可知：dx=(sect)^2dt,而由1+(tant)^2=(sect)^2知:
∫dx／(1+x²)= ∫(sect)^2dt/(sect)^2= ∫dt=t+C
而x=tant知t=arctanx即得出：
∫dx／(1+x²)= ∫(sect)^2dt/(sect)^2= ∫dt=t+C=arctanx+C