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

• lord12
In summary: I think I understand it now.In summary, the integral of dx/x^(2)-6x+82 can be solved using the substitution method and the trigonometric identity for integrals with a quadratic denominator. The final answer should involve tan^(-1) and can be simplified using the substitution u = x-3 and the value a = sqrt(73). There may be a discrepancy between different solutions, but using the given steps and notes, it can be resolved.
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?

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

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

What's the discrepancy?

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

lord12 said:
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

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

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

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

Last edited:
thanks people

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