Question

高数求不定积分

∫dx/(ax∧2+b)∧2 麻烦给个过程谢谢啦！

Answer 1

ax² + b = (√ax)² + (√b)² = (√btanz)² + (√b)² = b(tan²z + 1) = bsec²z
令√ax = √btanz，√a dx = √bsec²z dz
∫ dx/(ax² + b)²
= ∫ 1/(bsec²z)² * (√b/√a * sec²z dz)
= (√b/√a * 1/b²)∫ cos²z dz
= [1/(√ab^(3/2))](1/2)∫ (1 + cos2z)/2 dz
= [1/(2√ab^(3/2))](z + 1/2*sin2z) + C
= [1/(2√ab^(3/2))][arctan(√ax/√b) + √ax/√(ax² + b) * √b/√(ax² + b)] + C
= [1/(2√ab^(3/2))][arctan(√ax/√b) + √(ab)x/(ax² + b)] + C
= arctan(√a*x/√b)/[2√a*b^(3/2)] + x/[2b(ax² + b)] + C