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Question : 72 of 160
Marks:
+1,
-0
Solution:
Consider the expression,
∫dx =++ Now by partial fraction,
=+ + x=A(x−1)(x+2)+B(x+2) +C(x−1)2 Compare the coefficients,
A=,B=,C=− So,
∫dx=‌∫ +‌∫−‌∫ =‌log‌|x−1|−() −‌log‌|x+2|+c =−()+‌log||+c
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