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

[karush]

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...

 

[Prove It]

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 :)

 

[karush]

$$
\displaystyle
16\left[
-\frac{\ln\left({\left| x-2 \right|}\right)}{4}
+\frac{\ln\left({\left| x+2 \right|}\right)}{4} \right]
+\frac{{x}^{3}}{3}
+4x
$$
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

 

[Prove It]

As you originally negated everything in the first step, I think you need the negative of this as your answer.

 

[karush]

What was inside the brackets got turned around making the leading term positive

 

[Greg]

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. :)

 

[Prove It]

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

 

[karush]

$\textsf{Evaluate using decomposition}$
\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...

 

[Prove It]

$\textsf{Evaluate using decomposition}$
\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...

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

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