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Mathematics
Problem
∫
e
tan
−
1
x
(
1
+
x
+
x
2
1
+
x
2
)
d
x
\int e^{\tan^{-1}x}\left(\frac{1+x+x^2}{1+x^2}\right)\,dx
∫
e
t
a
n
−
1
x
(
1
+
x
2
1
+
x
+
x
2
)
d
x
is equal to (a)
x
e
tan
−
1
x
+
c
xe^{\tan^{-1}x} + c
x
e
t
a
n
−
1
x
+
c
(b)
x
2
e
tan
−
1
x
+
c
x^2e^{\tan^{-1}x} + c
x
2
e
t
a
n
−
1
x
+
c
(c)
1
x
e
tan
−
1
x
+
c
\frac{1}{x}e^{\tan^{-1}x} + c
x
1
e
t
a
n
−
1
x
+
c
(d) None of these (a)
x
e
tan
−
1
x
+
c
xe^{\tan^{-1}x} + c
x
e
t
a
n
−
1
x
+
c
(b)
x
2
e
tan
−
1
x
+
c
x^2e^{\tan^{-1}x} + c
x
2
e
t
a
n
−
1
x
+
c
(c)
1
x
e
tan
−
1
x
+
c
\frac{1}{x}e^{\tan^{-1}x} + c
x
1
e
t
a
n
−
1
x
+
c
(d) None of these
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