Find the indefinite integral.∫−2x2−12x+45dx\int_{}^{}\frac{-2}{x^2-12x+45}dx

− 2 3 t a n − 1 ( x − 6 3 ) + C -\frac23tan^{-1}\left(\frac{x-6}{3}\right)+C

Find the indefinite integral. ∫ − 2 x 2 − 12 x + 45 d x \int_{}^{}\frac{-2}{x^2-12x+45}dx

Here we're asked to find the indefinite integral of negative two over x squared minus 12 x plus 45 d x. Now looking at this here, if I were to choose a substitution where I set u equal to my denominator to try and get a natural log here, that's really not going to work because if I choose u as that x squared minus 12 x plus 45, that makes du equal to two x minus 12, which I definitely don't have in my integral. So when working with this here, I do want to go ahead and complete the square in this denominator so that I can get this in the form of an inverse trig function. So let's go ahead and do that here with this denominator. I have x squared minus 12 x. I wanna add some constant to that. I have my original constant 45 and then I'm going to subtract that same constant so that I'm not actually changing the value of this equation. Now the constant that we're adding and subtracting here is b over two squared. Here with a b value of negative 12, that's negative 12 over two squared, which is negative six squared, which is equal to 36. So we're adding 36 and we're also subtracting 36. This allows us to factor this down to x minus six squared. That's just that b value. Then we're adding that constant on the in there. 45 minus 36. That's just nine. So now I have rewritten my denominator here by completing the square. So now I can put that back into my original integral. Now I'm going to go ahead and keep this negative two out front using the constant multiple rule because I can already tell I am not going to need it when evaluating this interval. So I have that negative two out front. That's negative two times the integral of one over, and I'm going to go ahead and rewrite this in a slightly different order. Also rewriting that nine as three squared. So three squared plus X minus six squared. That's my new denominator there. So all I've done is rearrange this, put that in my denominator. So now I can see that this does take the form of one over a squared plus a U squared, where a is equal to three and u is equal to x minus six. Now, if u is x minus six, that makes du just equal to d x, which I have right there. So I can rewrite this as negative two times the integral of one over a squared plus u squared du. So I know that this will result in an inverse tangent. So this is gonna be negative two times one over a inverse tangent of u over a plus c. And we can of course plug in what u and a actually are here. So that makes this negative two over a. A is three, so that's negative two thirds times the inverse tangent of x minus six over three, plugging that u and a in, and then adding on our constant to the end there. So this is my final answer having evaluated that indefinite integral by completing the square in my integrand so that I could then use one of my inverse trig formulas. Let us know if you have any questions, and I'll see you in the next one.

Find the indefinite integral. ∫ − 2 x 2 − 12 x + 45 d x \int_{}^{}\frac{-2}{x^2-12x+45}dx

− 2 3 t a n − 1 ( x − 6 3 ) + C -\frac23tan^{-1}\left(\frac{x-6}{3}\right)+C

− 2 ln ⁡ ∣ x 2 − 12 x + 45 ∣ + C -2\ln\left|x^2-12x+45\right|+C

2 ln ⁡ ∣ x 2 − 12 x + 45 ∣ + C 2\ln\left|x^2-12x+45\right|+C

2 3 t a n − 1 ( x − 6 3 ) + C \frac23tan^{-1}\left(\frac{x-6}{3}\right)+C

11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals Involving Inverse Trigonometric Functions

Calculus

11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals Involving Inverse Trigonometric Functions

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Multiple Choice

Find the indefinite integral.
$\int_{}^{} \frac{- 2}{x^{2} - 12 x + 45} d x$

$- 2 \ln ∣ x^{2} - 12 x + 45 ∣ + C$

$2 \ln ∣ x^{2} - 12 x + 45 ∣ + C$

$\frac{2}{3} t a n^{- 1} (\frac{x - 6}{3}) + C$

$- \frac{2}{3} t a n^{- 1} (\frac{x - 6}{3}) + C$

Verified step by step guidance

Step 1: Recognize that the integral involves a rational function with a quadratic expression in the denominator. The denominator is x^2 - 12x + 45. To simplify, complete the square for the quadratic expression.

Step 2: Rewrite x^2 - 12x + 45 by completing the square. Factor out the coefficient of x^2 if necessary, then rewrite the quadratic as (x - 6)^2 - 9. This gives x^2 - 12x + 45 = (x - 6)^2 - 3^2.

Step 3: Substitute the completed square form into the integral. The integral becomes ∫(-2 / ((x - 6)^2 - 3^2)) dx. Recognize that this is a standard form for an integral involving an inverse tangent function.

Step 4: Use the standard formula for the integral of 1 / (u^2 - a^2), which is related to the inverse tangent function. Specifically, ∫(1 / (u^2 - a^2)) du = (1 / a) * arctan(u / a) + C. Adjust the constants and factor out -2 from the numerator.

Step 5: Perform the substitution u = x - 6 and a = 3 to match the standard form. Apply the formula to compute the integral, and include the constant of integration C. The final result will involve the arctan function with the appropriate coefficients.

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