∫\(\sqrt{\quad}\)(1+\(x^{2}\))dx の解法は
x+\(\sqrt{\quad}\)(1+\(x^{2}\))=tとおく
t-x=\(\sqrt{\quad}\)(1+\(x^{2}\))
\(t^{2}\)-2tx-1=0
x=\(\frac{t}{2}\)-1/(2t)

よって
\(\sqrt{\quad}\)(1+\(x^{2}\))=t-x=t-{\(\frac{t}{2}\)-1/(2t)}=\(\frac{t}{2}\)+1/(2t)=(\(t^{2}\)+1)/(2t)
dx={\(\frac{1}{2}\)+1/(2\(t^{2}\))}dt={(\(t^{2}\)+1)/(2\(t^{2}\))}dt
∫\(\sqrt{\quad}\)(1+\(x^{2}\))dx
=∫{(\(t^{2}\)＋1\()^{2}\)}/(4\(t^{3}\))dt
=(\(\frac{1}{4}\))∫(t+\(\frac{2}{t}\)+1/\(t^{3}\))dt
=\(t^{2}\)/8+(logt)/2-1/(8\(t^{2}\))
=(\(t^{4}\)-1)/(8\(t^{2}\))+(logt)/2

置き戻して(t=x+\(\sqrt{\quad}\)(\(x^{2}\)+1)でしたね)
{2\(x^{4}\)+2\(x^{2}\)+(2\(x^{3}\)+x)\(\sqrt{\quad}\)(\(x^{2}\)+1)}/{4\(x^{2}\)+2+4ｘx\(\sqrt{\quad}\)(\(x^{2}\)+1)}+log{x+\(\sqrt{\quad}\)(\(x^{2}\)+1)}/2
={\(x^{4}\)+\(x^{3}\)+\(x^{2}\)+x+(2\(x^{3}\)+x)\(\sqrt{\quad}\)(\(x^{2}\)+1)}/{4\(x^{2}\)+2+4x\(\sqrt{\quad}\)(\(x^{2}\)+1)}
=x\(\sqrt{\quad}\)(\(x^{2}\)+1)/2

よって
∫\(\sqrt{\quad}\)(1+\(x^{2}\))dx=x\(\sqrt{\quad}\)(\(x^{2}\)+1)/2+log{x+\(\sqrt{\quad}\)(\(x^{2}\)+1)}/2+C
Cは積分定数

x = tan(t), sin(t) = u
と置換して積分するのが一つの手か。答えは
(\(\frac{1}{2}\)){log(x+sqrt(1+\(x^{2}\)))+xsqrt(1+\(x^{2}\))}.

sinh(t)=(exp(t)-exp(-t))/2,cosh(t)=(exp(t)+exp(-t))/2とする。
(\(\frac{d}{d}\)t)sinh(t)=cosh(t),(\(\frac{d}{d}\)t)cosh(t)=sinh(t)
cosh(t\()^{2}\)-sinh(t\()^{2}\)=1となる。
∫\(\sqrt{\quad}\)(1+\(x^{2}\))dxにおいて、x=sinh(t)とする。dx=cosh(t)
積分は、∫{\(\sqrt{\quad}\)(1+sinh(t\()^{2}\))}cosh(t)dt
=∫cosh(t\()^{2}\)dt
=∫{exp(2t)+2+exp(-2t)}/4dt
=exp(2t)/2+\(\frac{t}{2}\)-exp(-2t)/2+C