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Question : 75 of 160
Marks:
+1,
-0
Solution:
Given,
∫(2x−3)√3x+2‌dxPut
‌‌3x+2=t ⇒‌‌3‌dx=dt ‌‌∫[2(‌)−3]√t‌dt =‌‌∫(2t−13)√t‌dt ⇒‌‌∫2t3∕2−13√t‌dt =‌[2⋅‌−13‌]+c =‌[‌t2−‌t]√t+c =‌[6(3x+2)2−65(3x+2)]√3x+2+c =‌[54x2−195x−106]√3x+2+c
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