Re: [微積] 六題積分題目 (需要算式)
※ 引述《milkbomb (牛奶炸彈客)》之銘言：
: 題目在這
: http://ppt.cc/gK1L
: 若是六題都會解可以站內信告知
: 解完後傳給我，給予酬勞臺幣200元謝之。謝謝
: (只需要一位)
1/2 1/4 2 3/4 1/2
A.∫(x + x ) dx = ∫(x + 2x + x )dx
2 7/4 3/2
x 8x 2x
= --- + --- + --- + C
2 7 3
2 1 2x -2x 2 1 4x -4x
B.[cosh(2x)] = ---(e + e ) = ---(e - e + 2)
4 4
4x -4x
2 1 4x -4x 1 e e
∫[cosh(2x)] dx = ---∫(e - e + 2)dx = ---(--- - --- + 2x) + C
4 4 4 4
x -x
e - e x -x x -x
C.∫tanhx dx = ∫------- dx, let u = e + e , du = (e - e )dx (by rule)
x -x
e + e
du x -x
∫tanhx dx = ∫-- = lnu + C = ln(e + e ) + C =ln(2coshx) + C
u
1 2 tan3x 1 2
D.∫(sec3x) e dx, let u = tan3x, du = ---(sec3x) dx
0 3
1 x=1 u 1 u|x=1 1 tan3x|1 1 tan3
---∫ e du = ---e | = ---e | = ---(e -1)
3 x=0 3 |x=0 3 |0 3
ln3
E.∫ (sech4x)(tanh4x)dx, let u = sech4x, du = -4(sech4x)(tanh4x) dx
ln2
x=ln3 -du -sech4x|ln3 1
∫ --- = -------| = ---[sech(4ln2) - sech(4ln3)]
x=ln2 4 4 |ln2 4
2
e 2t + 1 e 1 2 |e 2 2
F.∫ ------dt = ∫(2t + ---)dt = (t + lnt)| = e - 1 + lne +ln1 = e
1 t 1 t |1
--
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