利用级数求定积分的值∫(0到1)lnx*ln(1-x)dx
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Sievers分析仪
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总觉得这种瑕积分还是先求出原函数比较方便些。
∫
xln(1
-
x)
dx
=
∫
ln(1
-
x)
d(x²/2)
=
(x²/2)ln(1
-
x)
-
(1/2)∫
x²
*
(-
1)/(1
-
x)
dx
=
(x²/2)ln(1
-
x)
-
(1/2)∫
x²/(x
-
1)
dx
=
(x²/2)ln(1
-
x)
-
(1/2)∫
[(x²
-
1)
+
1]/(x
-
1)
dx
=
(x²/2)ln(1
-
x)
-
(1/2)∫
[(x
-
1)(x
+
1)
+
1]/(x
-
1)
dx
=
(x²/2)ln(1
-
x)
-
(1/2)∫
(x
+
1)
dx
-
(1/2)∫
dx/(x
-
1)
=
(x²/2)ln(1
-
x)
-
(1/2)(x²/2
+
x)
-
(1/2)ln|x
-
1|
+
c
=
(x²/2)ln(1
-
x)
-
x²/4
-
x/2
-
(1/2)ln|x
-
1|
+
c
=
(1/2)(x²
-
1)ln(1
-
x)
-
(x/4)(x
+
2)
+
c
∫(0→1)
xln(1
-
x)
dx
=
lim(x→1)
[(1/2)(x²
-
1)ln(1
-
x)
-
(x/4)(x
+
2)]
-
0
=
0
-
(1/4)(1
+
2)
=
-
3/4
∫
xln(1
-
x)
dx
=
∫
ln(1
-
x)
d(x²/2)
=
(x²/2)ln(1
-
x)
-
(1/2)∫
x²
*
(-
1)/(1
-
x)
dx
=
(x²/2)ln(1
-
x)
-
(1/2)∫
x²/(x
-
1)
dx
=
(x²/2)ln(1
-
x)
-
(1/2)∫
[(x²
-
1)
+
1]/(x
-
1)
dx
=
(x²/2)ln(1
-
x)
-
(1/2)∫
[(x
-
1)(x
+
1)
+
1]/(x
-
1)
dx
=
(x²/2)ln(1
-
x)
-
(1/2)∫
(x
+
1)
dx
-
(1/2)∫
dx/(x
-
1)
=
(x²/2)ln(1
-
x)
-
(1/2)(x²/2
+
x)
-
(1/2)ln|x
-
1|
+
c
=
(x²/2)ln(1
-
x)
-
x²/4
-
x/2
-
(1/2)ln|x
-
1|
+
c
=
(1/2)(x²
-
1)ln(1
-
x)
-
(x/4)(x
+
2)
+
c
∫(0→1)
xln(1
-
x)
dx
=
lim(x→1)
[(1/2)(x²
-
1)ln(1
-
x)
-
(x/4)(x
+
2)]
-
0
=
0
-
(1/4)(1
+
2)
=
-
3/4
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