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Question : 77 of 160
Marks:
+1,
-0
Solution:
Let
I=‌‌dxPut
1+√x=t‌⇒‌‌‌‌dx=dt‌⇒‌‌dx=2(t−1)‌dtWhen
x=0,t=1When
x=16,t=5‌∴I=‌t(2(t−1))‌dt‌‌‌=2‌‌‌=2‌(t−2+‌)‌dt‌‌‌=2[‌−2t+log‌t]15‌=2[{‌−10+log‌5}−{‌−2+log‌1}].
‌=2[‌+log‌5+‌+0]‌=8+2‌log‌5
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