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sesenic [268]
2 years ago
11

Evaluate the integral, show all steps please!

Mathematics
1 answer:
zalisa [80]2 years ago
4 0

Answer:

\dfrac{3}{2} \ln |x-4| - \dfrac{1}{2} \ln |x+2| + \text{C}

Step-by-step explanation:

<u>Fundamental Theorem of Calculus</u>

\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

\displaystyle \int \dfrac{x+5}{(x-4)(x+2)}\:\:\text{d}x

Take <u>partial fractions</u> of the given fraction by writing out the fraction as an <u>identity</u>:

\begin{aligned}\dfrac{x+5}{(x-4)(x+2)} & \equiv \dfrac{A}{x-4}+\dfrac{B}{x+2}\\\\\implies \dfrac{x+5}{(x-4)(x+2)} & \equiv \dfrac{A(x+2)}{(x-4)(x+2)}+\dfrac{B(x-4)}{(x-4)(x+2)}\\\\\implies x+5 & \equiv A(x+2)+B(x-4)\end{aligned}

Calculate the values of A and B using substitution:

\textsf{when }x=4 \implies 9 = A(6)+B(0) \implies A=\dfrac{3}{2}

\textsf{when }x=-2 \implies 3 = A(0)+B(-6) \implies B=-\dfrac{1}{2}

Substitute the found values of A and B:

\displaystyle \int \dfrac{x+5}{(x-4)(x+2)}\:\:\text{d}x = \int \dfrac{3}{2(x-4)}-\dfrac{1}{2(x+2)}\:\:\text{d}x

\boxed{\begin{minipage}{5 cm}\underline{Terms multiplied by constants}\\\\$\displaystyle \int ax^n\:\text{d}x=a \int x^n \:\text{d}x$\end{minipage}}

If the terms are multiplied by constants, take them outside the integral:

\implies \displaystyle \dfrac{3}{2} \int \dfrac{1}{x-4}- \dfrac{1}{2} \int \dfrac{1}{x+2}\:\:\text{d}x

\boxed{\begin{minipage}{5 cm}\underline{Integrating}\\\\$\displaystyle \int \dfrac{f'(x)}{f(x)}\:\text{d}x=\ln |f(x)| \:\:(+\text{C})$\end{minipage}}

\implies \dfrac{3}{2} \ln |x-4| - \dfrac{1}{2} \ln |x+2| + \text{C}

Learn more about integration here:

brainly.com/question/27805589

brainly.com/question/28155016

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<u>Step-by-step explanation:</u>

transform the parent graph of f(x) = ln x        into f(x) = - ln (x - 4)  by shifting the parent graph 4 units to the right and reflecting over the x-axis

(???, 0): 0 = - ln (x - 4)

            \frac{0}{-1} = \frac{-ln (x - 4)}{-1}

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            e^{0} = e^{ln (x - 4)}

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          <u> +4 </u>  <u>    +4 </u>

             5 = x

(5, 0)

(???, 1): 1 = - ln (x - 4)

            \frac{0}{-1} = \frac{-ln (x - 4)}{-1}

            1 = ln (x - 4)

            e^{1} = e^{ln (x - 4)}

             e = x - 4

          <u> +4 </u>   <u>    +4 </u>

         e + 4 = x

          6.72 = x

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Domain: x - 4 > 0

                <u>  +4 </u>  <u>+4  </u>

               x       > 4

(4, ∞)

Vertical asymptotes: there are no vertical asymptotes for the parent function and the transformation did not alter that

No vertical asymptotes

*************************************************************************

transform the parent graph of f(x) = 3ˣ        into f(x) = - 3ˣ⁺⁵  by shifting the parent graph 5 units to the left and reflecting over the x-axis

Domain: there is no restriction on x so domain is all real number

(-∞, ∞)

Range: there is a horizontal asymptote for the parent graph of y = 0 with range of y > 0.  the transformation is a reflection over the x-axis so the horizontal asymptote is the same (y = 0) but the range changed to y < 0.

(-∞, 0)

Y-intercept is when x = 0:

f(x) = - 3ˣ⁺⁵

      = - 3⁰⁺⁵

      = - 3⁵

      = -243

Horizontal Asymptote: y = 0  <em>(explanation above)</em>

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