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Question : 70 of 160
Marks:
+1,
-0
Solution:
Given,
∫tan−1(1−x+x2)dx+∫tan−1xdx+∫tan−1(1−x)dx =∫[−cot−1(1−x+x2)]dx+∫tan−1xdx+∫tan−1(1−x)dx =−∫tan−1()dx+∫tan−1xdx+∫tan−1(1−x)dx =−∫tan−1[]dx+∫6an−1xdx+∫tan−1(1−x)dx −∫[tan−1x+tan−1(1−x)−tan−1x−tan−1(1−x)]dx =x+C
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