Question

求不定积分∫x/√x²;+2x+3dx

Answer 1

∫ x/√(x² + 2x + 3) dx
= ∫ x/√[(x + 1)² + 2] dx
令x + 1 = √2 tanz，dx = √2 sec²z dz
=> ∫ (√2 tanz - 1)/√(2tan²z + 2) * (√2 sec²z) dz
= ∫ (√2 tanz - 1)/(√2 secz) * (√2 sec²z) dz
= ∫ (√2 tanz - 1) secz dz
= √2 ∫ secz tanz dz - ∫ secz dz
= √2 secz - ln|secz + tanz| + C
= √2 * √(x² + 2x + 3)/√2 - ln|√(x² + 2x + 3)/√2 + (x + 1)/√2| + C
= √(x² + 2x + 3) - ln|x + 1 + √(x² + 2x + 3)| + C
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tanz = (x+1)/√2
cosz = √2/√(x² + 2x + 3)