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Question : 42 of 115
Marks:
+1,
-0
Solution:
‌‌−‌‌‌−‌| 2()()2 |
| (1+(‌)2)√1+(‌)4 |
‌⇒‌−‌‌⇒‌‌dx‌⇒‌‌dx‌⇒−‌| 1− |
| (x+‌)√(x+‌)2−2 |
‌dx‌x+‌=t⇒1−‌‌dx=dt‌‌⇒−‌‌⇒−‌‌‌ take ‌t2−2=α2‌tdt=αdα‌⇒−‌‌⇒−‌‌⇒‌tan−1‌]∞√2‌⇒‌{tan−11}+‌tan−1∞‌⇒‌{‌−‌}‌⇒‌=‌‌‌ So ‌K=4√2‌K2=32
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