①　与式 =∫\(e^{x}\)/(e^(2x)+1) dx
t=\(e^{x}\) とおくと d\(\frac{t}{d}\)x=\(e^{x}\) より
　　　　　　=∫1/(\(t^{2}\)+1)dt
t=tanθ とおくと d\(\frac{t}{d}\)θ=\(\frac{1}{c}\)o\(s^{2}\)θ より
=∫co\(s^{2}\)θ ・\(\frac{1}{c}\)o\(s^{2}\)θ dθ
=∫dθ = θ +C
= tan^(-1)t +C
=tan^{-1}\(e^{x}\) + C

②　与式 = ∫\(x^{2}\)(sin x)' dx = \(x^{2}\)sin x-∫2xsin x dx
= \(x^{2}\) sin x +2∫x (cos x)' dx
= \(x^{2}\) sin x +2x cos x -2∫cos x dx
= \(x^{2}\) sin x + 2x cos x - 2sin x + C

(1) 被積分関数の分母分子に (\(e^{x}\)) を掛けて、\(e^{x}\) = \tan t と置けば、
\int 1/(\(e^{x}\)+e^{-x}) dx = \int dt = t + C = \arctan \(e^{x}\) + C.
(2) 部分積分を2回使って、
\int \(x^{2}\)\cos x dx = \(x^{2}\)\sin x - 2\int x\sin x dx
= \(x^{2}\)\sin x - 2(-x\cos x + \int\cos x dx)
= 2x\cos x + (\(x^{2}\)-2)\sin x + C.