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Question : 74 of 160
Marks:
+1,
-0
Solution:
Let
I=∫‌ Put,
x−2=‌dx=‌dt⇒x=2+‌ ∴‌‌I=−1∕t2‌| dt |
| √(2+‌)2−3(2+‌)+5 |
I=∫−‌=‌‌∫‌ I=‌sinh−1[‌] I=‌sinh−1[‌]+C[t=‌] I=‌sinh−1[‌]+C
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