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Today's Calculation Of Integral
2011 Today's Calculation Of Integral
736
736
Part of
2011 Today's Calculation Of Integral
Problems
(1)
Today's calculation of Integral 736
Source: created by kunny
7/3/2011
Evaluate
∫
0
1
(
e
x
+
1
)
{
e
x
+
1
+
(
1
+
x
+
e
x
)
ln
(
1
+
x
+
e
x
)
}
1
+
x
+
e
x
d
x
\int_0^1 \frac{(e^x+1)\{e^x+1+(1+x+e^x)\ln (1+x+e^x)\}}{1+x+e^x}\ dx
∫
0
1
1
+
x
+
e
x
(
e
x
+
1
)
{
e
x
+
1
+
(
1
+
x
+
e
x
)
ln
(
1
+
x
+
e
x
)}
d
x
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