How Do You Solve the Integral of dx/(x-3)² + 73?

[lord12]

integral of dx/x^(2)-6x +82

I know from completing the square you get the integral of dx/73 + (x-3)^(2). Then what?

 

[neutrino]

Do the substitution x-3 = u and see if it turns into something a little more familiar.

 

[lord12]

I just want to know what you guys think as there is a discrepancy between my professors answer and my answer.

 

[neutrino]

What's the discrepancy?

 

[lord12]

my professor says that the final answer involves tan^(-1). I say that its (1/73)(u^(-2)du) = (1/73)(-(x-3)^(-1).

 

[neutrino]

my professor says that the final answer involves tan^(-1). I say that its (1/73)(u^(-2)du) = (1/73)(-(x-3)^(-1).

Your professor is right. Check your work again, the denominator is u^2 + 73

 

[AlphaNumeric]

$$\int \frac{du}{u^{2}+A} = \int \frac{\sqrt{A}dy}{Ay^{2}+A} = \frac{1}{\sqrt{A}}\int \frac{dy}{y^{2}+1}$$

Combined with what others have said and what you should have in your notes you will be able to see the professor is right.

 

[courtrigrad]

$$\int \frac{dx}{x^{2}-6x+82} \;$$$$\int \frac{dx}{(x-3)^{2} + 73}$$Use the fact that $$\int \frac{dx}{x^{2}+a^{2}} = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C$$

So going back to the integral:

$$\int \frac{dx}{(x-3)^{2} + 73}$$, and $$u = x-3, \; du = dx$$

$$a^{2} = 73\Rightarrow a = \sqrt{73}$$

Thus $$\int \frac{dx}{(x-3)^{2} + 73} = \frac{1}{\sqrt{73}}\tan^{-1}\left(\frac{x-3}{\sqrt{73}}\right) + C$$

 

[lord12]

thanks people