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Question : 152 of 160
Marks:
+1,
-0
Solution:
Given that,
∫‌=f(x)+c . . . (i)
Let
‌‌I‌=∫‌‌=∫‌| dx |
| (√x+99)2+1)‌√x+99 |
Put
√x+99=t⇒‌‌dx=2‌dt∴‌‌I‌=∫‌=2tan−1t+c‌=2tan−1√x+99+cFrom Eqs. (i)
‌2tan−1√x+99+c=f(x)+c‌⇒‌‌f(x)=2tan−1√x+99‌
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