∫(3x+1)/[(√4+x²)] dx ∫sin√x dx

∫ (3x + 1)/√(4 + x²) dx
令x = 2tanθ,dx = 2sec²θ dθ
= ∫ (6tanθ + 1)/(2secθ) • (2sec²θ) dθ
= ∫ (6secθtanθ + secθ) dθ
= 6secθ + ln|secθ + tanθ| + C
= 6 • √(4 + x²)/2 + ln|√(4 + x²)/2 + x/2| + C
= 3√(4 + x²) + ln|x + √(4 + x²)| + C
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∫ sin√x dx
= ∫ sin√x • (2√x)/(2√x) dx
= 2∫ √xsin√x d(√x)
= -2∫ √x d(cos√x)
= -2√xcos√x + 2∫ cos√x d(√x)
= -2√xcos√x + 2sin√x + C