integral
no. 9
∫3/(2x - 2) dx
= 3/2 ∫1/(x - 1) d(x - 1)
= 3/2 ln (x - 1) + c
•
no. 10
∫6x/√(3x² + 2)
= ∫(3x² + 2)^(-1/2) d(3x² + 2)
= 1/(-1/2 + 1) (3x² + 2)^(-1/2 + 1) + C
= 2 (3x² + 2)^1/2 + C
= 2√(3x² + 2) + C
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Verified answer
integral
no. 9
∫3/(2x - 2) dx
= 3/2 ∫1/(x - 1) d(x - 1)
= 3/2 ln (x - 1) + c
•
no. 10
∫6x/√(3x² + 2)
= ∫(3x² + 2)^(-1/2) d(3x² + 2)
= 1/(-1/2 + 1) (3x² + 2)^(-1/2 + 1) + C
= 2 (3x² + 2)^1/2 + C
= 2√(3x² + 2) + C