Integration--what is the solution?

Question

Can some one help me with this solution .. thanks
integral of x/x^2+2x+5 (2,1)

so far I got ..... u-1/u^2+4

Answer 1

It's not clear what the integral is. You should always use parenthesis to avoid confusion. I guess you mean x/(x^2+2x+5). What does (2,1) mean?

Well, supposing that's the case, observe that

x/(x^2+2x+5) = (1/2) (2x + 2)/(x^2 + 2x + 5) -1/(x^2+2x+5)

Since 2x + 2 is the derivative of x^2 + 2x + 5, the integral of the 1st rational fraction is simply (1/2) ln(x^2+2x+5) + C1.

For the second, observe that x^2+2x+5 = (x +1)^2 + 2^2. We know that the integral of a function of the form du/(u^2 + a^2) is 1/a arctan(u/a). So, in our case, u = x+1, a =2 and du = dx. So, Int 1/(x^2+2x+5) = (1/2) arctan((x +1)/2) + C2.

Since C1 and C2 are constants, our final result is

(1/2) ln(x^2+2x+5) + C1 (1/2) arctan((x +1)/2) + C2 =
(1/2) [ln(x^2+2x+5) +arctan((x +1)/2) + C, where C = C1 + C2 is a constant.

If (2,1) means the integration limits, you just eveluate this last expression at x=2 and x =1 and subtract the second from the first

Answer 2

Integral of x/(x^2+2x+5) dx.

Note that the denominator can be rewritten into (x+1)^2 + 4 by completing the square.

Integral x/[(x+1)^2 + 4)] dx

Now make the substitution x+1 = u so that x = u-1 and dx = du,

Integral (u-1) / (u^2 + 4) du
= Integral u/(u^2+4) - 1/(u^2+4) du

The first part evaluates into 1/2 ln(u^2+4) while the second part evaluates into -1/2 arctan (u/2), so the answer is

1/2 ln (x^2+2x+5) - 1/2 arctan [(x+1)/2], evaluated at 2 and 1.

1/2 ln(13) - 1/2 arctan(3/2) - 1/2 ln(8) + 1/2 arctan(1)

= 1/2 [ln(13/8) - arctan(3/2) + arctan(1)]
or 1/2 [ln(13/8) - arctan(1/5)] if you use the tangent difference formula to combine the two arctans.