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Evaluation of definite integrals
Question
∫
0
1
tan
−
1
2
x
−
1
1
+
x
−
x
2
d
x
=
Moderate
A
π
4
B
ln
2
C
π
4
−
ln
2
D
0
Solution
I
=
∫
0
1
tan
−
1
2
x
−
1
1
+
x
−
x
2
d
x
,
x
→
1
−
x
I
=
∫
0
1
tan
−
1
1
−
2
x
1
+
x
−
x
2
d
x
2
I
=
0
→
I
=
0
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