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Mathematics
Problem
∫
e
2
x
+
log
x
d
x
=
\int e^{2x+\log x}\,dx =
∫
e
2
x
+
l
o
g
x
d
x
=
(a)
1
4
(
2
x
−
1
)
e
2
x
+
c
\frac{1}{4}(2x-1)e^{2x} + c
4
1
(
2
x
−
1
)
e
2
x
+
c
(b)
1
4
(
2
x
+
1
)
e
2
x
+
c
\frac{1}{4}(2x+1)e^{2x} + c
4
1
(
2
x
+
1
)
e
2
x
+
c
(c)
1
2
(
2
x
+
1
)
e
2
x
+
c
\frac{1}{2}(2x+1)e^{2x} + c
2
1
(
2
x
+
1
)
e
2
x
+
c
(d)
1
2
(
2
x
−
1
)
e
2
x
+
c
\frac{1}{2}(2x-1)e^{2x} + c
2
1
(
2
x
−
1
)
e
2
x
+
c
(a)
1
4
(
2
x
−
1
)
e
2
x
+
c
\frac{1}{4}(2x-1)e^{2x} + c
4
1
(
2
x
−
1
)
e
2
x
+
c
(b)
1
4
(
2
x
+
1
)
e
2
x
+
c
\frac{1}{4}(2x+1)e^{2x} + c
4
1
(
2
x
+
1
)
e
2
x
+
c
(c)
1
2
(
2
x
+
1
)
e
2
x
+
c
\frac{1}{2}(2x+1)e^{2x} + c
2
1
(
2
x
+
1
)
e
2
x
+
c
(d)
1
2
(
2
x
−
1
)
e
2
x
+
c
\frac{1}{2}(2x-1)e^{2x} + c
2
1
(
2
x
−
1
)
e
2
x
+
c
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