Question

求1/（cosx）4以及ln[x+（x2+1）1/2]的不定积分,

Answer 1

∫ 1/cos⁴x dx

= ∫ sec⁴x dx

= ∫ sec²x * sec²x dx

= ∫ (tan²x+1) dtanx

= (1/3)tan³x + tanx + C

∫ ln[x+√(x²+1)] dx

= x[x+√(x²+1)] - ∫ x dln[x+√(x²+1)]

= x[x+√(x²+1)] - ∫ x * 1/[x+√(x²+1)] * [1+x/√(x²+1)] dx

= x[x+√(x²+1)] - ∫ x/[x+√(x²+1)] * [√(x²+1)+x]/√(x²+1) dx

= x[x+√(x²+1)] - ∫ x/√(x²+1) dx

= x[x+√(x²+1)] - (1/2)∫ d(x²+1)/√(x²+1)

= x[x+√(x²+1)] - (1/2) * 2√(x²+1) + C

= x[x+√(x²+1)] - √(x²+1) + C