微分積分 II（ｃ） 補充問題１ 解答
問 1 (1) 9
(2) 1
6
(5) 1
2
(6) 1
1 cos 3x
3
1
(3) √ Tan−1 √x
3
3
(2) Sin−1 x
2
x−3
1
(4)
log 6
x+3
問 2 (1) −
問 3 (1)
1 x3 − 3 x2 + 3 log |x|
3
2
(3) −
(4) e2 − 1
(3) 2
(2) sin x + 2 tan x
x − √3 √
√ (4) x + 3 log x+ 3
1 −x
tan x
不定積分の置換積分
問 4 (1)
1 (x3 + 1)8
24
(4) Tan−1 (sin x)
1 (2x + 1)3
6
(4) 1 e2x
2
問 5 (1)
√
(3) − 1 (1 − x2 )3
3
2
(6) − 1 e−x
2
(2) 1 (log x)2
2
x
(5) Sin−1 ( e )
2
√
(2) 2 (3x − 1)3
9
(5) cos(1 − x)
1 log |x3 + 1|
3
(3) log | sin x|
問 6 (1)
(3) 1 log |5x + 2|
5
(6) 3 sin x
3
(2) 1 log |x2 + 2x − 3|
2
(4) log(ex + e−x )
不定積分の部分積分
問 7 (1) −xe−x − e−x (= −(x + 1)e−x )
問 8 (1) (−
問9 I =
1 x2 + 1 ) cos 2x + 1 x sin 2x
2
4
2
(2) 1 x sin 3x + 1 cos 3x
3
9
(2) x(log x)2 − 2x log x + 2x
1 e4x (4 cos 3x + 3 sin 3x)
25
定積分の置換積分
(2) e − 12
e
)
(
e
+
1
(4) log(e + 1) − log 2 = log
2
1
9
(5)
(6) 1 (e − 1)
log
3
2
2
問 10 (1) 10
(3) 1
4
定積分の部分積分
(
)
2
e−2
問 11 (1) − + 1 =
e
e
(
)
1 e2 + 1 = 1 (e2 + 1)
問 12
4
4
4
(2) π − 1
18
9
(
= π−2
18
)

Hs9 Exponentiële en logaritmische functies

Istant book 5 per mille 2015