Focusing on the I in I don't want to do this anymore

In summary: So $\displaystyle \begin{align*} \int{ \frac{x^4}{4 - x^2} \, \mathrm{d}x } &= -4\ln{ \left| x + 2 \right| } + 4\ln{ \left| x - 2 \right| } + \frac{x^3}{3} + 4\,x + C \end{align*}$and the answer is$\displaystyle \begin{align*} \int{ \frac{x^4}{4 - x^2} \, \mathrm{d}x } &= 4\ln{ \left| \frac{x - 2}{x +
  • #1
karush
Gold Member
MHB
3,269
5
w8.5.2
$$ \int \frac{x^4}{(4-x^2)} dx
\implies -\int\frac{x^4}{(x^2-4)} dx
$$
by division
$$\displaystyle\int\dfrac{16}{x^2-4}+x^2+4 \ dx
\implies 16\int \frac{1}{(x+2)(x-2)}+ x^2+4 \ dx $$
$$\frac{1}{(x+2)(x-2)}=\frac{A}{x+2}+\frac{B}{x-2}
\displaystyle = -\dfrac{1}{4\left(x+2\right)}+\dfrac{1}{4\left(x-2\right)}
$$
so now we have
$$-\frac{1}{4}\int\dfrac{1}{x+2} dx
\ \ + \ \frac{1}{4}\int\dfrac{1}{x-2} dx
\ \ +\int {x}^{2} \ dx
+ \int 4 \ dx $$
just seeing if going down the right trail...
 
Last edited:
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  • #2
karush said:
w8.5.2
$$ \int \frac{x^4}{(4-x^2)} dx
\implies -\int\frac{x^4}{(x^2-4)} dx
$$
by division
$$\displaystyle\int\dfrac{16}{x^2-4}+x^2+4 \ dx
\implies 16\int \frac{1}{(x+2)(x-2)}+ x^2+4 \ dx $$
$$\frac{1}{(x+2)(x-2)}=\frac{A}{x+2}+\frac{B}{x-2}
\displaystyle = -\dfrac{1}{4\left(x+2\right)}+\dfrac{1}{4\left(x-2\right)}
$$
so now we have
$$-\frac{1}{4}\int\dfrac{1}{x+2} dx
\ \ + \ \frac{1}{4}\int\dfrac{1}{x-2} dx
\ \ +\int {x}^{2} \ dx
+ \int 4 \ dx $$
just seeing if going down the right trail...

Yes it's fine, keep going :)
 
  • #3
$$
\displaystyle
16\left[
-\frac{\ln\left({\left| x-2 \right|}\right)}{4}
+\frac{\ln\left({\left| x+2 \right|}\right)}{4} \right]
+\frac{{x}^{3}}{3}
+4x
$$
Simplify
$$
\displaystyle
4\ln\left({\left| x+2 \right|}\right)
- 4\ln\left({\left| x-2 \right|}\right)
+\frac{{x}^{3}}{3}
+4x
+C
$$
I hope😍
 
  • #4
As you originally negated everything in the first step, I think you need the negative of this as your answer.
 
  • #5
What was inside the brackets got turned around making the leading term positive
 
  • #6
Prove It said:
Yes it's fine, keep going :)

No, it's not fine. The '16' is not applied to $x^2+4$ as the notation used implies. Sorry to nitpick but clean notation helps to avoid errors and improves marks. In my opinion it is an area were the OP could stand to make a substantial improvement. :)
 
  • #7
greg1313 said:
No, it's not fine. The '16' is not applied to $x^2+4$ as the notation used implies. Sorry to nitpick but clean notation helps to avoid errors and improves marks. In my opinion it is an area were the OP could stand to make a substantial improvement. :)

I'm not sure why an eight month old thread has been bumped in the first place, I'm sure in that time the OP has already completed the problem, and probably the course it was from :P
 
  • #8
$\textsf{Evaluate using decomposition}$
\begin{align}
I_{02}&=\int \frac{x^4}{(4-x^2)} dx \\
&=-4\int \frac{1}{x+2} \, dx +4\int\frac{1}{x-2}+\int x^2 \, dx + 4\int 1 \, dx \\
\\
&=-4\ln{|x+2|}+4\ln{|x-2|}+\frac{x^3}{3}+4x+C
\end{align}

How dis...:cool:
 
  • #9
karush said:
$\textsf{Evaluate using decomposition}$
\begin{align}
I_{02}&=\int \frac{x^4}{(4-x^2)} dx \\
&=-4\int \frac{1}{x+2} \, dx +4\int\frac{1}{x-2}+\int x^2 \, dx + 4\int 1 \, dx \\
\\
&=-4\ln{|x+2|}+4\ln{|x-2|}+\frac{x^3}{3}+4x+C
\end{align}

How dis...:cool:

Let's just end this...

$\displaystyle \begin{align*} \int{ \frac{x^4}{4 - x^2} \, \mathrm{d}x } &= -\int{ \frac{x^4}{x^2 - 4} \, \mathrm{d}x } \\ &= -\int{ \frac{ x^4 - 4\,x^2 + 4\,x^2 }{ x^2 - 4 } \, \mathrm{d}x } \\ &= -\int{ \left( \frac{ x^4 - 4\,x^2 }{ x^2 - 4 } + \frac{ 4\,x^2 }{ x^2 - 4 } \right) \,\mathrm{d}x } \\ &= -\int{ \left[ \frac{ x^2 \, \left( x^2 - 4 \right) }{x^2 - 4} + \frac{4\,x^2}{ x^2 - 4 } \right] \, \mathrm{d}x } \\ &= - \int{ \left( x^2 + \frac{4\,x^2}{x^2 - 4} \right) \,\mathrm{d}x } \\ &= -\int{ \left( x^2 + \frac{4\,x^2 - 16 + 16}{ x^2 - 4} \right) \,\mathrm{d}x } \\ &= -\int{ \left( x^2 + \frac{4\,x^2 - 16}{x^2 - 4} + \frac{16}{x^2 - 4} \right) \,\mathrm{d}x } \\ &= -\int{ \left[ x^2 + \frac{4\,\left( x^2 - 4 \right) }{x^2 - 4} + \frac{16}{x^2 - 4} \right] \,\mathrm{d}x } \\ &= -\int{ \left( x^2 + 4 + \frac{16}{x^2 - 4} \right) \,\mathrm{d}x } \\ &= -\int{ \left[ x^2 + 4 + \frac{16}{\left( x - 2 \right) \left( x + 2 \right) } \right] \,\mathrm{d}x } \end{align*}$

Now applying partial fractions:

$\displaystyle \begin{align*} \frac{A}{x - 2} + \frac{B}{x + 2} &\equiv \frac{16}{\left( x - 2 \right) \left( x + 2 \right) } \\ A \,\left( x + 2 \right) + B \,\left( x - 2 \right) &\equiv 16 \end{align*}$

Let $\displaystyle \begin{align*} x = -2 \end{align*}$ to find $\displaystyle \begin{align*} -4\,B = 16 \implies B = -4 \end{align*}$.

Let $\displaystyle \begin{align*} x = 2 \end{align*}$ to find $\displaystyle \begin{align*} 4\,A = 16 \implies A = 4 \end{align*}$, giving

$\displaystyle \begin{align*} -\int{ \left[ x^2 + 4 + \frac{16}{ \left( x - 2 \right) \left( x + 2 \right) } \right] \,\mathrm{d}x } &= -\int{ \left( x^2 + 4 + \frac{4}{x - 2} - \frac{4}{x + 2} \right) \,\mathrm{d}x } \\ &= - \left( \frac{x^3}{3} + 4\,x + 4\ln{ \left| x - 2 \right| } - 4\ln{ \left| x + 2 \right| } \right) + C \\ &= 4\ln{ \left| x + 2 \right| } - 4\ln{ \left| x - 2 \right| } - \frac{x^3}{3} - 4\,x + C \end{align*}$
 

FAQ: Focusing on the I in I don't want to do this anymore

What is W8.5.1 integral expansion?

W8.5.1 integral expansion is a mathematical method for finding the antiderivative of a function. It is also known as the method of undetermined coefficients.

How does W8.5.1 integral expansion work?

W8.5.1 integral expansion involves breaking down a function into simpler terms and then using a table of known integrals to find the antiderivative. This method is particularly useful for functions that cannot be easily integrated using basic integration rules.

What is the purpose of W8.5.1 integral expansion?

The purpose of W8.5.1 integral expansion is to find the antiderivative of a function, which is important in many areas of science and engineering. It is also used to solve differential equations and in the process of integration by parts.

Are there any limitations to W8.5.1 integral expansion?

Yes, W8.5.1 integral expansion may not work for all functions. It is most effective for functions that can be broken down into simpler terms using algebraic manipulation. It may also result in a complex antiderivative in some cases.

Can W8.5.1 integral expansion be used for definite integrals?

Yes, W8.5.1 integral expansion can be used to find the definite integral of a function. After finding the antiderivative using this method, the lower and upper limits of integration can be substituted into the function to find the definite integral.

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