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Byju's Answer
Standard XII
Physics
Antiderivative
If ∫01tan-1x-...
Question
If
1
∫
0
tan
−
1
x
cot
−
1
(
1
−
x
+
x
2
)
d
x
equals
k
,
then the value of
1
k
is
A
3
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B
2
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C
7
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D
13
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Solution
The correct option is
B
2
k
=
1
∫
0
tan
−
1
x
cot
−
1
(
1
−
x
+
x
2
)
d
x
…
(
1
)
⇒
k
=
1
∫
0
tan
−
1
x
cot
−
1
(
(
1
−
x
)
2
+
x
)
d
x
Applying
b
∫
a
f
(
x
)
d
x
=
b
∫
a
f
(
a
+
b
−
x
)
d
x
,
k
=
1
∫
0
tan
−
1
(
1
−
x
)
cot
−
1
(
(
1
−
1
+
x
)
2
+
(
1
−
x
)
)
d
x
⇒
k
=
1
∫
0
tan
−
1
(
1
−
x
)
cot
−
1
(
x
2
+
1
−
x
)
d
x
…
(
2
)
From
(
1
)
+
(
2
)
,
2
k
=
1
∫
0
tan
−
1
(
1
−
x
)
+
tan
−
1
(
x
)
cot
−
1
(
x
2
+
1
−
x
)
d
x
⇒
2
k
=
1
∫
0
tan
−
1
(
x
+
1
−
x
1
+
x
(
1
−
x
)
)
cot
−
1
(
x
2
+
1
−
x
)
d
x
⇒
2
k
=
1
∫
0
tan
−
1
(
1
x
2
+
1
−
x
)
cot
−
1
(
x
2
+
1
−
x
)
d
x
⇒
2
k
=
1
∫
0
1
d
x
⇒
k
=
1
2
∴
1
k
=
2
Suggest Corrections
0
Similar questions
Q.
If
I
=
∫
1
0
cot
−
1
(
1
−
x
+
x
2
)
d
x
=
k
∫
1
0
tan
−
1
x
d
x
,
then
k
equals
Q.
If
I
=
∫
1
0
cot
−
1
(
1
−
x
+
x
2
)
dx
=
k
∫
1
0
tan
−
1
x
dx
, then the value of k equals