Question

xln(1-x)dx定积分 下限0 上限1 .求定积分的值有过程有真相

xln(1-x)dx定积分 下限0 上限1 .求定积分的值 有过程有真相

Answer 1

简单计算一下即可，答案如图所示

Answer 2

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