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Question : 59 of 160
Marks:
+1,
-0
Solution:
etI=∫‌dx=∫‌dx Put
‌‌xn=t differentiate w.r.to '
x '
nxn−1dx=dt‌‌⇒‌‌xn−1dx=‌ ∴‌‌I=f‌=‌‌∫‌dt =‌⋅‌tan−1(‌)+c [∵∫‌dx=‌tan−1(‌)+c] =‌tan−1(‌)+c
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