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Question : 76 of 160
Marks:
+1,
-0
Solution:
‌‌ Let ‌I=‌‌dx‌=‌| (x2−1)(x2+1) |
| (x2−1)(x4+x2+1) |
‌dx‌=‌‌dx=(‌)‌dx‌=(‌)‌dx‌=‌‌dxLet
x−‌=t⇒(1+‌)‌dx=dtWhen
x⟶1, then
t⟶0When
x⟶2, then
t⟶‌I‌=‌‌dt‌=‌[tan−1‌]03∕2‌=‌tan−1‌‌=‌tan−1‌
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