Question

Evaluate the integral, show all steps please!

Answer 1

Answer:

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

Step-by-step explanation:

Fundamental Theorem of Calculus

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

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 partial fractions of the given fraction by writing out the fraction as an identity:

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

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:

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

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

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

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Answer 2

For an alternative approach, expand and complete the square in the denominator to write

$(x-4)(x+2) = x^2 - 2x - 8 = (x - 1)^2 - 9$

In the integral, substitute $x - 1 = 3 \sin(u)$ and $dx=3\cos(u)\,du$ to transform it to

$\displaystyle \int \frac{x+5}{(x - 1)^2 - 9} \, dx = \int \frac{3\sin(u) + 6}{9 \sin^2(u) - 9} 3\cos(u) \, du \\\\ ~~~~~~~~~~~~ = - \int \frac{\sin(u) + 2}{\cos(u)} \, du \\\\ ~~~~~~~~~~~~ = - \int (\tan(u) + 2 \sec(u)) \, du$

Using the known antiderivatives

$\displaystyle \int \tan(x) \, dx = - \ln|\cos(x)| + C$

$\displaystyle \int \sec(x) \, dx = \ln|\sec(x) + \tan(x)| + C$

we get

$\displaystyle \int \frac{x+5}{(x - 1)^2 - 9} \, dx = \ln|\cos(u)| - 2 \ln|\sec(u) + \tan(u)| + C \\\\ ~~~~~~~~~~~~ = - \ln\left|\frac{(\sec(u) + \tan(u))^2}{\cos(u)}\right|$

Now, for $n\in\Bbb Z$,

$\sin(u) = \dfrac{x-1}3 \implies u = \sin^{-1}\left(\dfrac{x-1}3\right) + 2n\pi$

so that

$\cos(u) = \sqrt{1 - \dfrac{(x-1)^2}9} = \dfrac{\sqrt{-(x-4)(x+2)}}3 \implies \sec(u) = \dfrac3{\sqrt{-(x-4)(x+2)}}$

and

$\tan(u) = \dfrac{\sin(u)}{\cos(u)} = -\dfrac{x-1}{\sqrt{-(x-4)(x+2)}}$

Then the antiderivative we found is equivalent to

$\displaystyle - \int \frac{x+5}{(x - 1)^2 - 9} \, dx = - \ln\left|-\frac{3(x+2)}{(x-4) \sqrt{-(x-4)(x+2)}}\right| + C$

and can be expanded as

$\displaystyle - \int \frac{x+5}{(x - 1)^2 - 9} \, dx = -\ln\left| \frac{3(x+2)^{1/2}}{(x-4)^{3/2}}\right| + C \\\\ ~~~~~~~~~~~~ = - \ln\left|(x+2)^{1/2}\right| + \ln\left|(x-4)^{3/2}\right| + C \\\\ ~~~~~~~~~~~~ = \boxed{\frac32 \ln|x-4| - \frac12 \ln|x+2| + C}$