Find the integral of ;

(X^2-29x+5)÷{(x-4)(x^2+3)}

[Dyuman Joshi]

Never thought I'd see a day when someone named Newton would ask a calculus question, which is unrelated to the forum on which the question is asked.

[Walter Roberson]

The integral over what range? The symbolic integral has 3 distinct ranges (two discontinuities) and getting the phase right can be difficult.

Answer 1

[Sam Chak]

Hello Sir @Newton

I feel honored to have the opportunity to solve this integral problem for you in the modern computational era.

% Plotting the rational function
x1 = -30:0.00001:4;
x2 =   4:0.00001:30;
f  = @(x) (x.^2 - 29*x + 5)./((x - 4).*(x.^2 + 3));
plot(x1, f(x1), 'color', '#0d6ea3'), hold on
plot(x2, f(x2), 'color', '#0d6ea3'), ylim([-10 10])
xline(4, '-.r', 'x = 4'), grid on
xlabel('x'), ylabel('f(x)')

% defining the expression
syms x real
expr = (x^2 - 29*x + 5)/((x - 4)*(x^2 + 3))

expr = 

% performing partial fraction decomposition 
expr = partfrac(expr, x)

expr = 

% finding the indefinite integral
F    = int(expr)

F = 

[Sam Chak]

For the first term, we can factor out the constant

and then substitute so that :

Then, find the integral formula from the Schaum's Outline of Mathematical Handbook of Formulas and Tables to get

For the second term, find the integral formula from the Handbook to obtain

Multiply the numerator by yields

.

The last term can be solved by taking the substitution approach like the first term.

[Walter Roberson]

syms x LB UB real
assumeAlso(LB <= UB)
expr = (x^2 - 29*x + 5)/((x - 4)*(x^2 + 3))

expr = 

int(expr, x, LB, UB)

ans = 

So the definite integral cannot be found if the bounds cross 4.

[David Goodmanson]

Actually the integral can be found even if the bounds cross 4, assuming the principal value is taken. From both Walter's result and SC's partial fraction expresson it's clear that the only part of the integrand that involves crossing 4 is

-5/(x-4)

whose itegral is

-5* [ log(UB-4) -log(LB-4) ]

If LB and UB are on opposite sides of 4, then the principal value is

-5* lim{eps-->0)  [  Int{LB, 4-eps} 1/(x-4) dx + Int{4+eps, UB} 1/(x-4) dx ]

with the only consequence being that absolute values appear in the log terms:

-5* [ (log(abs(UB-4)) -log(abs(LB-4)) ]

If the principal value is always implied, the expression just above works no matter where UB, LB are. Same side, opposite sides, both positive, both negative, doesn't matter.

[Walter Roberson]

Testing to see what int() can do.

I find it interesting that taking the integral and substituting in bounds, gets a different representation of the result than if you take the integral supplying the bounds -- different enough that all my tests with rewrite() and simplify() were not able to get val2s - val4s to collapse to 0.

syms x LB UB real
assumeAlso(LB <= UB)
expr = (x^2 - 29*x + 5)/((x - 4)*(x^2 + 3))

expr = 

assumeAlso(LB < 4)
assumeAlso(4 < UB)

val1 = int(expr, x, LB, UB, 'PrincipalValue', false)

val1 = 

NaN

val2 = int(expr, x, LB, UB, 'PrincipalValue', true)

val2 = 

val2s = simplify(subs(val2, [LB, UB], [3 5]), 'steps', 10)

val2s = 

val3 = int(expr, x, 3, 5, 'PrincipalValue', false)

val3 = 

NaN

val4 = int(expr, x, 3, 5, 'PrincipalValue', true)

val4 = 

val4s = simplify(val4, 'steps', 10)

val4s = 

val2s - val4s

ans = 

simplify(ans, 'steps', 25)

ans = 

vpa(ans)

ans = 

3.6734198463196484624023016788195e-40