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Question : 76 of 160
Marks:
+1,
-0
Solution:
∵x2+x+1=(x+‌)2+(‌)2 ∴∫√x2+x+1dx=∫√(x+‌)2+(‌)2dx =‌√x2+x+1+‌‌ln|(x+‌)+√x2+x+1| =‌√x2+x+1+‌‌ln|(‌)+√x2+x+1| and
∫‌dx=∫‌dx =‌‌log|(x+‌)+√x2+x+1| =‌‌log|‌+√x2+x+1| ‌‌ Now, ‌∫√x2+x+1dx×∫‌dx =‌‌[‌√x2+x+1+‌‌log|(‌)+√x2+x+1|] ×[log(‌+√x2+x+1)] =(‌√x2+x+1+‌sinh−1‌) (sinh−1‌)+C
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