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Mathematics
Problem
∫
e
tan
−
1
x
(
1
+
x
+
x
2
)
⋅
d
(
cot
−
1
x
)
\int e^{\tan^{-1} x} (1+x+x^2) \cdot d(\cot^{-1} x)
∫
e
t
a
n
−
1
x
(
1
+
x
+
x
2
)
⋅
d
(
cot
−
1
x
)
is equal to - (A)
−
e
tan
−
1
x
+
c
-e^{\tan^{-1} x} + c
−
e
t
a
n
−
1
x
+
c
(B)
e
tan
−
1
x
+
c
e^{\tan^{-1} x} + c
e
t
a
n
−
1
x
+
c
(C)
−
x
⋅
e
tan
−
1
x
+
c
-x \cdot e^{\tan^{-1} x} + c
−
x
⋅
e
t
a
n
−
1
x
+
c
(D)
x
⋅
e
tan
−
1
x
+
c
x \cdot e^{\tan^{-1} x} + c
x
⋅
e
t
a
n
−
1
x
+
c
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