Evaluate the integral. (Remember to use absolute values where appropriate. Remember the constant of integration.) J 13 150 13 125 dx 2 ln(|x − 5| ) − In ( x² + 5x+25|) + 2√3 tan 2x -1 ( ²53 +/- 5 ) + c) 5√3

College Algebra
1st Edition
ISBN:9781938168383
Author:Jay Abramson
Publisher:Jay Abramson
Chapter5: Polynomial And Rational Functions
Section: Chapter Questions
Problem 27PT: Find the unknown value. 27. y varies jointly with x and the cube root of 2. If when x=2 and...
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Hello, I am working paritial fraction decompsoition for integrals and I am not sure where I am going wrong on this problem. Attached is my work and the problem.

3= B+ 4D
2013 dx f
dis
x²2²-125
23
(X-5) (x²+57 +25)
S&x=-5
43 (6+6)= (6-6) (6² +66 +6²)
26
- 1x15 dx
245725
9
Bf dr
ts #fl d² - 1/² ( +x+²) dr
5
45+25
A= 1/3
dx-2
B=-13
75
13= A (x² + 5x+25) + (Bxt() (x-5) C = -26
13 - A (25+25 + 25) +e
11/²² = A
13
13
yufde #fcez de
dx
X-5
75 25 +25
4 2 1 x 5) - 12/23 (2) 21
X
3
A
X-5
25
BX+C
13 = A6 ² + 5/²² +25A + 8x.2-15 Betfr 52
13= (A+B) ₂ ² + (5A-5B+C) x + (25A-52)
C=A+B -> C= 12 +13
B=-13
13 = 25/12) =52
13=13 56
5
26=-56-36²-26²
(x+ 2)²+25
h=x+²
+5x125
55
GE
6
泰
— TAN (2)
TE
1/2 In/x-5) - 12/3 (2 In 1945 ×+25) - 5 5,TAN" (245)) + C
4₂
²+5+25
X+2
Sa
75
J(²²5) ² 250
4= x+²₂2 + u-Zcx
dorito
→>
X+2
(24150125) -25 +25
S
4-5/+2
R
S
4²+75
M
dr
du →
do os fis to do
・du
75
Sand & Su de Susdor - S w
6425
with
drtu
4 { lolute 2 - 2 ( 3 TON 1 (24))
4₂ = 10 |(x + 5) ²+ 25 - 5TAN (2 (5)
AM
553
TAN
I = 40/1n | x ²²5x + ²5 + ²1 - S (165)
503
Transcribed Image Text:3= B+ 4D 2013 dx f dis x²2²-125 23 (X-5) (x²+57 +25) S&x=-5 43 (6+6)= (6-6) (6² +66 +6²) 26 - 1x15 dx 245725 9 Bf dr ts #fl d² - 1/² ( +x+²) dr 5 45+25 A= 1/3 dx-2 B=-13 75 13= A (x² + 5x+25) + (Bxt() (x-5) C = -26 13 - A (25+25 + 25) +e 11/²² = A 13 13 yufde #fcez de dx X-5 75 25 +25 4 2 1 x 5) - 12/23 (2) 21 X 3 A X-5 25 BX+C 13 = A6 ² + 5/²² +25A + 8x.2-15 Betfr 52 13= (A+B) ₂ ² + (5A-5B+C) x + (25A-52) C=A+B -> C= 12 +13 B=-13 13 = 25/12) =52 13=13 56 5 26=-56-36²-26² (x+ 2)²+25 h=x+² +5x125 55 GE 6 泰 — TAN (2) TE 1/2 In/x-5) - 12/3 (2 In 1945 ×+25) - 5 5,TAN" (245)) + C 4₂ ²+5+25 X+2 Sa 75 J(²²5) ² 250 4= x+²₂2 + u-Zcx dorito →> X+2 (24150125) -25 +25 S 4-5/+2 R S 4²+75 M dr du → do os fis to do ・du 75 Sand & Su de Susdor - S w 6425 with drtu 4 { lolute 2 - 2 ( 3 TON 1 (24)) 4₂ = 10 |(x + 5) ²+ 25 - 5TAN (2 (5) AM 553 TAN I = 40/1n | x ²²5x + ²5 + ²1 - S (165) 503
Evaluate the integral. (Remember to use absolute values where appropriate. Remember the constant of integration.)
1 =
13
x³ - 125
dx
13
2x +
150 (2 m ( x − 5) - In ( x² + 5x + 25 ) + 2√J tan¯¹ ( 25 ) + C)
5√3
X
Transcribed Image Text:Evaluate the integral. (Remember to use absolute values where appropriate. Remember the constant of integration.) 1 = 13 x³ - 125 dx 13 2x + 150 (2 m ( x − 5) - In ( x² + 5x + 25 ) + 2√J tan¯¹ ( 25 ) + C) 5√3 X
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