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Question : 72 of 160
Marks:
+1,
-0
Solution:
I=∫ =∫ =∫ =∫ Put
u= ⇒du=| (1+x2)(2x)−(2+x2)(2x) |
| (1+x2)2 |
dx. ⇒du=| 2x(1+x2−2−x2)dx |
| (1+x2)2 |
⇒du= ⇒=− Then,
I=−‌∫ =−[]+C =×(u)+C =()+C Thus,
α= and
n=5 Then,
==2
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