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Question : 76 of 160
Marks:
+1,
-0
Solution:
I=∫(x+2)√x+3dx Put
x+3=t2 ⇒x‌‌=t2−3 dx‌‌=2tdt ∴I‌‌=∫(t2−1)⋅2t2dt ‌‌=2(‌−‌)+C=‌(3t5−5t3)+C ‌‌=‌[3⋅(x+3)2√x+3−5(x+3)√x+3]+C ‌‌=‌√x+3[3(x2+6x+9)−5(x+3)+C ‌‌=‌√x+3(3x2+13x+12)+C
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