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MHT CET 2017 Math Solved Paper
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© examsnet.com
Question : 5
Total: 50
1
∫
0
x
tan
−
1
x
d
x
=
π
4
+
1
2
π
4
−
1
2
1
2
−
π
4
−
π
4
−
1
2
Validate
Solution:
(b)
1
∫
0
x
tan
−
1
x
d
x
=
=
[
tan
−
1
x
∫
x
d
x
]
0
1
-
1
∫
0
(
d
d
x
(
tan
−
1
x
)
∫
x
d
x
)
d
x
=
[
tan
−
1
x
.
x
2
2
]
0
1
-
1
∫
0
(
1
1
+
x
2
⋅
x
2
2
)
d
x
.
=
(
1
2
tan
−
1
1
−
0
)
-
1
2
1
∫
0
1
+
x
2
−
1
1
+
x
2
d
x
=
1
2
(
π
4
)
-
1
2
1
∫
0
(
1
−
1
1
+
x
2
)
d
x
=
π
8
−
1
2
[
x
−
tan
−
1
x
]
0
1
=
π
8
−
1
2
[
1
−
tan
−
1
1
−
0
+
0
]
=
π
8
−
1
2
[
1
−
π
4
]
=
π
4
−
1
2
© examsnet.com
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