1.
∫1/cos(x)dx
=
∫cosx*dx/(cos(x))^2
=
∫d(sinx)/(1-(sinx)^2)
=-1/2*∫2*d(sinx)/((sinx-1)*(sinx+1)
=-1/2*((∫d(sinx)/(sinx-1 ) - ∫d(sinx)/(sinx+1))
=1/2*ln((1+sinx)/(1-sinx))+C
2.
已知
(x√(1+x*x))'
=√(1+x*x) + x*x/√(1+x*x)
=√(1+x*x) + ((x*x+1)-1)/√(1+x*x)
=2√(1+x*x) - 1/√(1+x*x)
關(guān)鍵是求1/√(1+x*x)的原函數(shù)
設(shè)
x=tany,y屬于(-pai/2,pai/2)
∫1/√(1+x*x)
=∫1/cosy dy
=1/2*ln((1-siny)/(1+siny))+C
=ln(x + √(1+x*x))+C
=>
∫√(1+x*x)dx
=1/2*(x√(1+x*x)+ln(x+√(1+x*x))+C