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Today's Calculation Of Integral
2011 Today's Calculation Of Integral
723
723
Part of
2011 Today's Calculation Of Integral
Problems
(1)
Today's calculation of Integral 723
Source: created by kunny
6/27/2011
Evaluate
∫
1
e
{
1
−
(
x
−
1
)
e
x
}
ln
x
(
1
+
e
x
)
2
d
x
.
\int_1^e \frac{\{1-(x-1)e^{x}\}\ln x}{(1+e^x)^2}dx.
∫
1
e
(
1
+
e
x
)
2
{
1
−
(
x
−
1
)
e
x
}
l
n
x
d
x
.
calculus
integration
logarithms
calculus computations