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Question : 74 of 160
Marks:
+1,
-0
Solution:
Let
I=∫‌put
x2=t⇒2x‌dx=dt⇒x‌dx=‌∴I‌=∫‌I‌=‌‌∫‌=‌‌∫‌Now,
‌=‌+‌⇒‌‌t=A(t+2)+B(t+1)When,
t+2=0⇒t=−2∴‌‌−2=B(−1)⇒B=2when,
t+1=0⇒t=−1‌∴‌‌−1=A(1)+B(0)⇒A=−1‌∴‌=‌+‌‌∴I=‌‌∫(‌+‌)‌dt‌I=‌[−log(t+1)+2‌log(t+2)]+C‌=−‌‌log(x2+1)+log(x2+2)]+C‌=log(x2+2)+log(x2+1)−1∕2+C‌=log(‌)+C
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