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Question : 44 of 160
Marks:
+1,
-0
Solution:
Consider the given integral
∫‌‌=∫‌‌dx=∫‌‌dx−∫‌‌dx =∫‌‌dx−∫‌‌dx =log|x|−∫‌‌dx+C Let,
x4+1=t After differentiation this gives
4x3dx=dt x3dx=‌‌dt Substitute these values, we get,
∫‌‌=log|x|−‌‌‌∫‌‌+C =log|x|−‌‌‌log|x4+1|+C =‌‌[log|x4|−log|x4+1|]+C =‌‌‌log|‌‌|+C
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