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Question : 144 of 150
Marks:
+1,
-0
Solution:
cot−1(1−x+x2)dx =tan−1() =tan−1() =tan−1() ⇒cot−1(1−x+x2)=tan−1x −tan−1(1−x) ∴cot−1(1−x+x2)dx=tan−1xdx−tan−1(1−x)dx =tan−1xdx+tan−1xdx[∵f(x)dx=−f(a−x)dx] =2‌tan−1xdx On evaluating by integration by parts, we have
=2{[tan−1x.x]01−dx} =2{−[‌ln(1+x2)]01} =2[−‌log‌2]=−log‌2 Hence,
cot−1(1−x+x2)dx=−log‌2 So, option (b) is correct.
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