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I've been stuck on this problem for 2 hours and I have no idea where I'm going wrong. Here is the original equation:

$$\int_{0}^{1}\frac{x-4}{x^2-5x+6}dx$$

I first factored and rewrote the fraction this way:

$$\frac{A(x-1)+B(x-6)}{(x-6)(x+1)}$$

After expanding and grouping I got this:

$$\frac{(A+B)x+(A-6B)}{(x-6)(x+1)}$$

I then solved the system of equations by means of algebraic substitution and got $A=\frac{2}{7}$ and $B=\frac{5}{7}$.

I rewrote my integral using my new $A$ and $B$ values:

$$\int_{0}^{1}\frac{2}{7x-42}+\frac{5}{7x+7}dx$$

I tried integrating the above by factoring out a $\frac{1}{7}$ from both terms and I tried integrating it in unfactored form. Neither answer was correct.

In factored form I got the following:

$$\frac{1}{7}(2ln|x-6|)+\frac{1}{7}(5ln|x+1|)|_{0}^{1}$$

In unfactored form I got the following:

$$14ln|7x-42|+35ln|7x+7||_{0}^{1}$$

Is everything up to this point correct and my error with the evaluation of the final step or did I make a mistake earlier on?

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    $\begingroup$ You factored incorrectly $x^2-5x+6 = (x-2)(x-3)$ $\endgroup$ – Amzoti Sep 28 '13 at 1:29
  • $\begingroup$ @Amzoti omg, you're right. How embarrassing! $\endgroup$ – hax0r_n_code Sep 28 '13 at 1:30
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    $\begingroup$ @Amzoti thank you for the encouragement. Calculus 2 is the hardest class I've ever taken and I refuse to give up. This place has helped and inspired me so much. Thanks for the giving of your time to help people! $\endgroup$ – hax0r_n_code Sep 28 '13 at 1:32
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    $\begingroup$ We have all had classes where we felt like that. Practice, patience and persistence! You are doing fine, keep at it and that is why most of us participate on the site! Give a yell if you get stuck, but it will probably be smooth going after that fix. Regards $\endgroup$ – Amzoti Sep 28 '13 at 1:37
  • $\begingroup$ I was looking at this problem for a while until I scrolled down to the comments. Thanks, Amzoti. $\endgroup$ – user153085 Nov 5 '14 at 1:29
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You factored incorrectly.

It should be:

$$ x^2−5x+6=(x−2)(x−3)$$

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