高数不定积分习例题讲解(二)
高 等 数 学(1)学 习 辅 导(9)
不定积分习例题讲解(二)
计算题
1. 已知f ' (x ) =sec 2x +sin x ,且f (0) =1,求函数f (x ) 解:f (x ) =⎰f ' (x ) dx =⎰(sec2x +sin x ) dx
=⎰sec 2xdx +⎰sin xdx =tan x -cos x +c
将f (0) =1代入,得c =2 则f (x ) =tan x -cos x +2 2. 若⎰f (t ) dt =F (t ) +c , 则
⎰
f (at +b ) dt =
1
F (at +b ) +c 1 (a ≠0) a
证明:
用不定积分定义证明:
⎰f (t ) dt =F (t ) +c , 所以F ' (t ) =f (t )
又
d 11
[F (at +b )]=F ' (u ) u =at +b (at +b )' dt a a
=F ' (u ) u =at +b =f (u ) u =at +b =f (at +b )
1
∴F (at +b ) 是f (at +b ) 的一个原函数, a
1
由不定积分定义有⎰f (at +b ) dt =F (at +b ) +c 1
a
另证,用第一换元积分法证明。
1
f (at +b ) dt =f (at +b ) d (at +b ) ⎰⎰a
11
f (u ) du =[F (u ) +c ] a ⎰a 1
=F (at +b ) +c 1 a
采用“凑微分,使变量一致”的方法,将所求积分化成已知结果的不定积分或基本积分公式的那种形式,就可求得该积分。
=
3.
解:
x 3
dx 4.⎰2
1+x
解:
x 31x 21(1+x 2) -122
⎰1+x 2dx =2⎰1+x 2dx =2⎰1+x 2(x )
11122=[⎰dx 2-⎰d (x +1) ]=[x -ln(x 2+1)]+c 2
221+x
5.
解:
⎰ 6.
x 1-x
2
d x
解:利用第一换元法
⎰
x -x 2
d x =⎰
12-x 2
d(x 2) =-⎰
12-x 2
d(1-x 2)
=-⎰d(-x 2) =-x 2+c
7.
解:
8.
解:
3
9.. tan xdx
⎰
32tan xdx =tan ⎰⎰x . tan xdx
=⎰(sec2x -1) tan xdx =⎰sec 2x tan xdx -⎰tan xdx
=⎰tan xd (tanx ) -⎰=
sin x
dx cos x
11tan 2x +⎰d (cosx ) 2cos x 1
=tan 2x +ln cos x +c 2
10.
解:
-x 2
11. ⎰d x 2
x
解: 利用第二换元法,设x =sin t ,d x =cos t d t
⎰
-x 2cos t ⋅cos t
d x =⎰sin 2t d t x 2
2
1-s i n t 1
d t =(-1) t d =⎰22⎰s i n s i n t t
-x 2
=-cot t -t +c =-+arcsin x +c
x
12.
解:
d x 13. ⎰a r c s i x n
解:利用分部积分法
⎰arcsin x d x =x arcsin x -⎰x d (arcsinx )
i n -⎰ =x a r c s x
=x arcsin x +⎰
x -x 12-x 2
2
d x
d(1-x 2)
=x a r c s x i n +-x 2+c
14.
解:
15.求不定积分
⎰
x 2+a 2dx
x
, dx =a sec 2tdt , a
解:第二换元法求解
令x =a tan t , t =arctan
⎰x 2+a 2dx =⎰a 2sec 3tdt .
32
又 sec tdt =sec t . sec tdt =sec td tan t
⎰⎰⎰
=sec t . tan t -⎰tan 2t sec tdt =sec t . tan t -⎰(sec2t -1) sec tdt =sec t . tan t +⎰sec tdt -⎰sec 3tdt
x 223
. x +a +ln sec t +tan t -sec tdt 2⎰a x 1
∴⎰sec 3tdt =2x 2+a 2+ln x +x 2+a 2+c
22a
=
则
⎰
x a 222
x +a dx =. x +a +ln x +x 2+a 2+c
22
2
2
另解(分部积分法求解)
⎰x 2+a 2dx =x x 2+a 2-⎰xd (x 2+a 2)
=x x +a -⎰=x x +a -⎰
2
2
22
x 2x +a
2
2
dx
(x 2+a 2) -a 2
x +a
2
2
dx
1x +a
2
2
=x x 2+a 2-⎰x 2+a 2dx +a 2⎰
dx
=x x 2+a 2+a 2ln x +x 2+a 2-⎰x 2+a 2dx
则
⎰
x 2a 22
x +a dx =x +a +ln x +x 2+a 2+c
22
2
2
16.
ln x
⎰x 2d x
解: 利用分部积分法
ln x 1ln x 1
d x =ln x d(-) =-+⎰x 2⎰⎰x d(ln x ) x x
ln x 1ln x 1
+⎰2d x =--+c =-x x x x
17.
(n ≠-1)
解:
18. 求积分xf ' ' (x ) dx 解:
⎰
⎰xf ' ' (x ) dx =⎰xd [f ' (x )]=xf ' (x ) -⎰f ' (x ) dx
=xf ' (x ) -f (x ) +c
19.
解:
20.
解:
21.求积分
⎰
1x +a
2
2
dx
2
解:令x =a tan t , dx =a sec tdt
⎰
1x 2+a 2
dx =⎰
1
a sec 2tdt =⎰sec tdt a sec t
=ln sec t +tan t +c 1
x 2+a 2x =ln ++c 1
a a
=ln x +x 2+a 2+c , 其中c =c 1-ln a