Calc 2 Partial Fractions... why is this wrong?

Question

Evaluate:
Integral( (2x^2 + 5) / (x^2 + 1)(x^2 + 4) dx)
Set it up with partial fractions...
(Ax + B)/(x^2 + 1) + (Cx + D)/(x^2 + 4)
Multiply through by denom..
Ax(x^2 + 4) + B(x^2 + 4) + Cx(x^2 + 1) + D(x^2 + 1) = 2x^2  +5
Multiply out..
Ax^3 + 4Ax + Bx^2 + 4B + Cx^3 + Cx + Dx^2 + D
Solve for variables..
4Ax + Cx = 0x
4B + D = 5
A + C = 0
B + D = 2
4A + C = 0
4B + D = 5
A = -C
4 * -C + C = 0
-3C = 0
C = 0
A = 0
D = 2 - B
4B + 2 - B = 5
5B = 3
B = 3/5
D = 7/5

Plug variables back in..
(0x + 3/5) / (x^2 + 1) + (0x + 7/5) / (x^2 + 4)
3/5Integral(1/(x^2 + 1) dx) + 7/5Integral(1/(x^2 + 4) dx)
Integrate..
3/5[arctan(x)] + 7/5[1/2arctan(x/2)]
3/5arctan(x) + 7/10 arctan(x/2) + C

This is not the correct answer.
Why?
What is the correct answer?
Does Calc 2 ever stop being impossible?

? · Accepted Answer

4B + 2 - B = 5
5B = 3  ERROR 
B = 3/5
D = 7/5

3B=3
B=1
D=1

= 1/(x^2 + 1) + 1/(x^2 + 4)

Anonymous · Answer

I got
A=C=0
B=D=1

Anonymous · Answer

I presume that the x before everything is develop into self sustaining from some thing so as which you're fairly attracted to a attributable to combine 4/(x^2 + x). that's separated into partial fractions via skill of first noting that x^2 + x = x(x + one million). then you chop up 4/x(x + one million) suited right into a/x + b/(x + one million). you hit upon the values of a and b via skill of putting all of it decrease back over the classic denominator of x(x + one million) to offer you trouble-free simultaneous equations for a and b. If the appropriate replaced into (x + 4) the attitude is rather a matching. attempt it your self first and upload a be conscious as quickly as you have authentic situation.

Calc 2 Partial Fractions... why is this wrong?

3 Answers