x=tant,dx=sec²tdt
∫dx/[(2x^2+1)(x^2+1)^(1/2) ]
=∫sec²tdt/[(2tan²t+1)sect]
=∫dt/[cost((2sin²t/cos²t)+1)]
=∫costdt/[((2sin²t+cost²)]
=∫[1/(1+sin²t)]d(sint)
=arctan(sint)+C
三角替換有sint=x/√(1+x²)
所以原不定積分
∫dx/(2x^2+1)(x^2+1)^(1/2)
=arctan[x/√(1+x²)]+C