求1/（cosx）4以及ln[x+（x2+1）1/2]的不定积分,
2个回答
∫ 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
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