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Question : 152 of 160
Marks:
+1,
-0
Solution:
Let
I=∫x49‌‌dxLet
x50=t⇒50x49‌dx=dt∴‌‌I=‌‌∫‌‌dtLet
tan−1t=u⇒‌‌dt=du∴‌‌I‌=‌‌∫udu=‌+c‌=‌+c But
I=k(tan−1x50)2+c (given)
⇒k(tan−1x50)2+c‌=‌(tan−1x50)2+c⇒k=‌
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