Integral of (3x^2+1)/(3x^3+x^2+x+1) using partial fractions

Question

Please answer with step by step solution. Need ASAP. Thanks!

Answer 1

Rolls up the sleeves, not for the weak at heart......

the irrational solution is approximately -0.635

Synthetic division says:

-0.635 | 3 1 1 1

-1.905 .574635 -1

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3 -.905 1.574635 0

the quotient is

3x^2 - 0.905x + 1.574635

As stated, this polynomial is APPROXIMATE!!!

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PFD:

A/ (x+ 0.635) + (Bx+C) / (3x^2 - 0.905x + 1.574635) = left side

Multiplies by LCD, the right side becomes:

A(3x^2 - 0.905x + 1.574635) + (Bx+C)(x+ 0.635)

Distributes A and FOILS right hand term

3A x^2 - 0.905A x + 1.574635A + Bx^2 + 0.635Bx + Cx + 0.635C

CLT:

(3A+B)x^2 + (0.635B + C - 0.905A)x + 1.574635A + 0.635C = 3x^2 + 0x + 1

Equating coefficients:

3A + B = 3

(0.635B + C - 0.905A) = 0

1.574635A + 0.635C = 1

The system has solution:

A=.65784 B = 1.02647925274152757455 C = -.056469342

Check:

.65784 (3x^2 - 0.905x + 1.574635) + (1.02647925274152757455x-.056469342)(x+ 0.635) =

1.97352 x^2 - .5953452 x + 1.0358578884 + 1.02647925274152757455x^2 + .65181432549087000983925x

-.056469342x - 0.03585803217

the coefficient of x^2 is close to 3

the coefficient of x is close to zer0

the tail constant is close to 1

so A,B, and C are GOOD !!!!

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the first term integrates to A ln | x+ 0.635 |

Integrating the second term:

(Bx+C) / (3x^2 - 0.905x + 1.574635) =

(Bx + C) / 3(x^2 - 0.905/3 x + 1.574635/3)

multiplies the denominator by 2/b*B/2:

(Bx + C) / 3*2/b (b/2)(x^2 - 0.905/3 x + 1.574635/3)

(Bx+C) / (6/b) [ (b/2)(x^2) - (0.905*b/6)x + 1.574635B/6 ]

Note that -(0.905*b/6) is approximately C ... we'll call it k

Let U = [ (b/2)(x^2) +kx + 1.574635B/6 ]

dU = bx + k

the integrand becomes:

dU/U

and 6/b in the denominator factors out

the second term then approximately integrates to:

B/6 ln | U |

You can finish it from there

I agree that this integral should be calculated numerically using either SIMPSON's rule of 3/8 or

some other numerical integration, as we are approximating what is already an approximation.