>>344
>∫√(X^2+1)dxがどうやったらx√(X^2+1)+log(x+√(X^2+1))

t=√(x^2+1) +x　で置換。
ただし、(1/2) (x√(x^2+1)+log(x+√(x^2+1) )のはず。

dt=(x/√(x^2+1) +1 ) dx、x^2=(t^2-1)^2/(4t^2)より、
∫(x^2+1)/(√(x^2+1) +x ) dt=∫(t/4 +1/(2t) +1/(4t~3) )dt
=(1/2)logt +1/8(t^2-1/t^2)
=(1/2)log(x+√x^2+1)+(1/2)(x√(x^2+1))