Split the function in exercises into partial fractions(a) and find the antideriv

Question

Split the function in exercises into partial fractions(a) and find the antiderivative of the function in that exercise(b)

1) 20/[25-x(^2)]

2) [x+1]/[6x+x(^2)]

3) 2(1-s)/ s[s(^2)+3s+2]

4) 1/[w(^4)-w(^3)]

Evaluate the integral

1)  integral [(dx)/(x(^3)-x(^2)]  use [A/x] + [B/(x^2)]  + [C/(x-1)]

2)  integral {[x(^4)+12X(^3) +15x(^2)+25x+11]/ [x(^3)+12x(^2)+11x]} dx

     use division and (A/x)+(B/x+1)+(C/x+11)

Complete the square and give a substitution which could be used to compute the integrals in problem.

1) integral  [(x+1) / x(^2)+2x+ 2]dx

Explanation / Answer

Split the function in exercises into partial fractions(a) and find the antiderivative of the function in that exercise(b)
1) 20/[25-x(^2)]

20/(5-x)(5+x) = A/(5-x) +B/(5+x)
20 = A(5+x) +B(5-x)
at x = 5
20 = 10A => A = 20/10 = 2
at x = -5
20 = 10B => B = 20/10 = 2
20/(25-x^2) = 2/(5-x) + 2/(5+x)
taking integration we get
-2 ln(5-x) + 2ln(5+x) + C => 2ln((5+x)/(5-x))+ C

2) [x+1]/[6x+x(^2)]

(x+1)/x(6+x)) = A/x + B/(6+x)
x+1 = A(6+x) + Bx
at x = 0
1 = 6A => A = 1/6
at x = -6
-6+1 = -6B => B = -5/-6 = 5/6
(x+1) /(6x+x^2) = 1/6*1/x +5/6*1/(6+x) taking integration we get
= 1/6 ln(x) + 5/6 ln(x+6) + C

3) 2(1-s)/ s[s(^2)+3s+2]
2(1-s)/ s[s(^2)+3s+2]
2(1-s)/(s(s+1)(s+2)) = A/s +B/(s+1)+ C/(s+2)
2(1-s) = A(s+1)(s+2) + Bs(s+2) + C(s)(s+1)
at s = 0
2*1 = A*1*2 => 2A =2 => A = 2/2 = 1
at s = -1
2*(1+1) = 2*2 = B*-1*1 = > B = -4
at s = -2
2*3 = C-2*-1 = > 2C = 6 => C = 6/2 = 3
2(1-s)/(s(s+1)(s+2)) = 1/s -4/(s+1) + 3/(s+2) taking integration
= ln(s) -4ln(s+1) + 3ln(s+2) + C

4) 1/[w(^4)-w(^3)]
1/[w(^4)-w(^3)]
1/(w^3(w-1) = A/w + B/w^2+C/w^3+D/(w-1)
1 = Aw^2(w-1) +Bw(w-1) + C(w-1) + Dw^3
at w = 0
-C = 1 => C = -1
at w = 1
D = 1
A+D = 0 => A = -D => A = -1 (by comparing coefficeints of w^3 on both sides)
B-A = 0 => B = A => B = -1    (by comparing coefficeints of w^2 on both sides)
1/(w^3(w-1) = -1/w -1/w^2-1/w^3+1/(w-1) taking integration
= - ln(w) +1/w +1/2x^2 + ln(w-1) + C

Evaluate the integral
1) integral [(dx)/(x(^3)-x(^2)] use [A/x] + [B/(x^2)] + [C/(x-1)]
1/(x^2(x-1) = A/x + B/x^2 +C/(x-1)
1 = Ax(x-1)+B(x-1) + C x^2
1= Ax^2-Ax+Bx-B + Cx^2
1 = (A+C)x^2 + (B-A)x -B comparing coefficients we get...
A+C = 0
B-A =0
-B = 1 => B = -1
A = B   => A = -1
C = -A => C = 1
1/(x^2(x-1) = -1/x - 1/x^2 +1/(x-1) taking integration we get
= -ln(x) + 1/x + ln(x-1) + C

2) integral {[x(^4)+12X(^3) +15x(^2)+25x+11]/ [x(^3)+12x(^2)+11x]} dx
use division and (A/x)+(B/x+1)+(C/x+11)
[x(^4)+12X(^3) +15x(^2)+25x+11]/ [x(^3)+12x(^2)+11x]}
(x^4+12X^3 +15x^2+25x+11)/ (x^3+12x^2+11x)

x^3+12x^2+11x) x^4 + 12x^3 + 15x^2+ 25x+11 ( x
                x^4 + 12x^3 + 11x^2
               --------------------------------
                              4x^2+25x+11
                          
(x^4+12X^3 +15x^2+25x+11)/ (x^3+12x^2+11x) = x + (4x^2+25x+11) /(x(x^2+12x+11) = x+ (4x^2+25x+11)/(x(x+11)(x+1)
(4x^2+25x+11)/(x(x+1)(x+11) = A/x +B/(x+1)+ C/(x+11)
(4x^2+25x+11) = A(x+1)(x+11) + Bx(x+11) + Cx(x+1)
4x^2+25x+11 = A(x^2+12x+11) +B(x^2+11x) +C(x^2+x)
4x^2+25x+11 = (A+B+C)x^2+ (12A+11B+C)x + (11A)
comparing coefficeints 11A = 11 => A =1
A+B+C = 4 => B+C = 4-1 = 3
12A+11B+C = 25 => 11B+C = 25-12 = 13
11B+C =13
B+C = 3
10B = 10 => B = 1
C = 3-1 = 2
(x^4+12X^3 +15x^2+25x+11)/ (x^3+12x^2+11x) = x + 1/x + 1/(x+1) + 2/(x+11) taking integration we get
x^2/2 + ln(x) + ln(x+1)+ 2ln(x+11) + C

Complete the square and give a substitution which could be used to compute the integrals in problem.
1) integral [(x+1) / x(^2)+2x+ 2]dx
x^2+2x+2 = x^2+2x+1 +1 = (x+1)^2 + 1
(x+1)/((x+1)^2+1) dx let (x+1)^2 = t => 2(x+1) dx = dt
(x+1) dx = dt/2 integrand will be
dt/2*1/(1+t^2) = 1/2 dt/(1+t^2)   taking integration we get
= tan^-1(t)/2 + C
= tan^-1((x+1)^2)) + C

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