How do you integrate (x-1)/(x^2+4x+5)?

Question

Partial Fractions?

Answer 1

The user who answered before me didn't read the question carefully. His answer is wrong.

To integrate this, observe that

(x-1)/(x²+4x+5) = 1/2 (2x + 4)/(x²+4x+5) - 3/(x²+4x+5)  = 1/2 (2x + 4)/(x²+4x+5) -3/[(x +2)² + 3] = 1/2 (2x + 4)/(x²+4x+5) -3/[(x +2)² + (&radic3)²]

Since (x²+4x+5)' = 2x + 4, ∫1/2 (2x + 4)/(x²+4x+5) dx = 1/2 ln|x^2+4x+5| + C1

We know that ∫ du/(u² + a²) = 1/a arctan(x/a) + C. Hence,

∫3/[(x +2)² + 3] dx = 3/√3 arctan((x + 2)/√3) + C2

Finally, ∫(x-1)/(x²+4x+5) dx = 1/2 ln|x^2+4x+5| - √3 arctan((x + 2)/√3) + C

EDIT: There's a mistake in Gerald's answer, (x² + 4x + 5) does NOT equal (x+5)(x-1). The roots of  x² + 4x + 5 are -2 + i and -2 -i. Hence, it can't be factored over the reals.

Answer 2

∫ (x-1) / (x² + 4x + 5) dx

factorize the denominator (x² + 4x + 5) = (x+5)(x-1)

= ∫ (x-1) / [(x+5)(x-1)] dx

cancel out (x-1)

= ∫ 1 / (x+5) dx

= ln(x+5) + c   &bnsp; &bnsp; {answer}

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Here is the detail for the answer

∫ (x-1) / (x² + 4x + 5) dx

factorize the denominator (x² + 4x + 5) = (x+5)(x-1)

= ∫ (x-1) / [(x+5)(x-1)] dx

work now with the integrand only:

the factors of the denominator is linear, so use the following procedure

break the denominator apart into fractions

(x-1) / [(x+5)(x-1)] = A/(x+5) + B/(x-1)

then multiply (A) with the denominator of (B)

and multiply (B) with the denominator of (A)

in order to add them like regular fractions, to "restore" the denominator of the integrand

A/(x+5) + B/(x-1) = [A(x-1) + B(x+5)] / [(x+5)(x-1)]

set the "new" fraction and main integrand equal, then the denominators cancels out, since both sides have the same denominator

(x-1) / [(x+5)(x-1)] = [A(x-1) + B(x+5)] / [(x+5)(x-1)]

denominators cancels out

x - 1 = A(x-1) + B(x+5)

next, find the values for (x) so that (A) and (B) can be zero

now, if (x=1) then [A(x-1) = 0]

and, if (x=-5) then [B(x+5) = 0]

if (x=1) then (B=0)

x - 1 = A(x-1) + B(x+5)

1 - 1 = A(1-1) + B(1+5)

0 = 0 + B6

B = 0/6

B = 0

if (x = -5) then (A=1)

x - 1 = A(x-1) + B(x+5)

-5 - 1 = A(-5 - 1) + B(-5 + 5)

-6 = A(-6) + 0

A = -6 / -6

A = 1

return back to the main integral, then break up and substitute (A = 1) and (B = 0)

∫ (x-1) / (x² + 4x + 5) dx

= ∫ A/(x+5) + B/(x-1) dx

= ∫ 1/(x+5) + 0/(x-1) dx

= ∫ 1/(x+5) dx + ∫ 0/(x-1) dx

= 1∙ln(x+5) + 0∙ln(x-1) + c

= ln(x+5) + 0 + c

= ln(x+5)

ANSWER

= ln(x+5) + c

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I hope this helps

be good