Can't factor the denominator in this integral

uberbandgeek6 · #1 $Old$ September 27th, 2009, 06:33 PM

Problem: find the integral of (x-1) / (x^2 - 4x + 5)

I read somewhere that it was a good idea to complete the square, but that just left me with the integral of (x - 1) / ( (x-2)^2 + 9 ) , which doesn't seem to be much of a help. What is the best way to approach this problem?

Chris L T521 · #2 $Old$ September 27th, 2009, 06:39 PM

Quote:

Originally Posted by uberbandgeek6 $View Post$

Problem: find the integral of (x-1) / (x^2 - 4x + 5)

I read somewhere that it was a good idea to complete the square, but that just left me with the integral of (x - 1) / ( (x-2)^2 + 9 ) , which doesn't seem to be much of a help. What is the best way to approach this problem?

When you complete the square, you're left with

$\int\frac{x-1}{(x-2)^2+1}\,dx$ .

Now, if you want to integrate this, note that $\int\frac{x-1}{(x-2)^2+1}\,dx=\int\frac{x-2+1}{(x-2)^2+1}\,dx=\int\frac{x-2}{(x-2)^2+1}\,dx+\int\frac{\,dx}{(x-2)^2+1}$

For $\int\frac{x-2}{(x-2)^2+1}\,dx$ , use the substitution $u=(x-2)^2+1$

For $\int\frac{\,dx}{(x-2)^2+1}$ , use the substitution $u=x-2$ .

Can you take it from here?

DeMath · #3 $Old$ September 27th, 2009, 06:44 PM

Quote:

Originally Posted by uberbandgeek6 $View Post$

Problem: find the integral of (x-1) / (x^2 - 4x + 5)

I read somewhere that it was a good idea to complete the square, but that just left me with the integral of (x - 1) / ( (x-2)^2 + 9 ) , which doesn't seem to be much of a help. What is the best way to approach this problem?

$\int {\frac{{x - 1}} {{{x^2} - 4x + 5}}dx} = \frac{1} {2}\int {\frac{{2x - 4 + 2}} {{{x^2} - 4x + 5}}dx} =$

$= \frac{1} {2}\int {\frac{{2x - 4}} {{{x^2} - 4x + 5}}dx} + \int {\frac{{dx}} {{{x^2} - 4x + 5}}} .$

$\frac{1} {2}\int {\frac{{2x - 4}} {{{x^2} - 4x + 5}}dx} = \left\{ \begin{gathered} {x^2} - 4x + 5 = u, \hfill \\ \left( {2x - 4} \right)dx = du \hfill \\ \end{gathered} \right\} =$

$= \frac{1} {2}\int {\frac{{du}} {u}} = \frac{1} {2}\ln \left| u \right| + C = \frac{1} {2}\ln \left( {{x^2} - 4x + 5} \right) + C.$

$\int {\frac{{dx}} {{{x^2} - 4x + 5}}} = \int {\frac{{dx}} {{{{\left( {x - 2} \right)}^2} + 1}}} = \left\{ \begin{gathered} x - 2 = u, \hfill \\ dx = du \hfill \\ \end{gathered} \right\} =$

$= \int {\frac{{dx}}{{{u^2} + 1}}} = \arctan u + C = \arctan \left( {x - 2} \right) + C.$

Finally, you have

$\boxed{\int {\frac{{x - 1}}{{{x^2} - 4x + 5}}dx} = \frac{1}{2}\ln \left( {{x^2} - 4x + 5} \right) + \arctan \left( {x - 2} \right) + C}$

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O! Chris L T521 has already done.