Question

这高数题不会写？

Answer 1

(5)
let
√x = 3tanu
dx/(2√x) = 3(secu)^2 du
dx =18tanu.(secu)^2 du
∫√(x+9)/x dx
=∫{ 3secu/[9(tanu)^2] } .[18tanu.(secu)^2 du]
=6∫ (secu)^3/tanu du
=6∫ du/[ sinu. (cosu)^2 ]
=6∫ sinu/[ (sinu)^2. (cosu)^2 ] du
=-6∫ dcosu/[ (sinu)^2. (cosu)^2 ]
=-6∫ dcosu/[ (1-cosu)(1+cosu)(cosu)^2 ]
=-6∫ { (1/2)[1/(1-cosu)] +(1/2)[1/(1+cosu)] +1/(cosu)^2 } dcosu
=∫ { -3[1/(1-cosu)] -3[1/(1+cosu)] -6/(cosu)^2 } dcosu
=3ln|1-cosu| -3ln|1+cosu| +6/cosu + C
=3ln| 1-3/√(x+9)| -3ln|1+3/√(x+9)| + 6/[3/√(x+9)] + C
=3ln| √(x+9) -3| -3ln|√(x+9)+3 | + 2√(x+9) + C
where
√x = 3tanu
tanu =√x/3
cosu = 3/√(x+9)
let
1/[(1-cosu)(1+cosu)(cosu)^2] ≡ A/(1-cosu) +B/(1+cosu) +C/cosu +D/(cosu)^2
=>
1≡ A(1+cosu)(cosu)^2 +B(1-cosu)(cosu)^2 +C(1-cosu)(1+cosu)(cosu)
+D(1-cosu)(1+cosu)
cosu =0, => D=1
cosu=1, => A = 1/2
cosu=-1, => B=1/2
coef. of (cosu)^3
A-B-C=0
1/2-1/2 -C =0
C=0
ie
1/[(1-cosu)(1+cosu)(cosu)^2] ≡ (1/2)[1/(1-cosu)] +(1/2)[1/(1+cosu)] +1/(cosu)^2
(6)
let
√x =tanu
dx/(2√x) = (secu)^2 du
dx=2tanu. (secu)^2 du
∫ dx/[x.√(x+1)]
=∫ 2tanu. (secu)^2 du/[ (tanu)^2.secu]
=2∫ (secu/tanu) du
=2∫ cscu du
=2ln|cscu-cotu| +C
=2ln|√[(x+1)/x] - 1/√x | + C
=2ln|√(x+1) -1 | -ln|x| + C
where
√x =tanu
sinu = √[x/(x+1)]
cscu =1/sinu =√[(x+1)/x]
cotu =1/√x

Answer 2

题目发过来！