# Thread: w8.5.1 integral expansion

1. 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....

2. Originally Posted by 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....
Yes it's fine, keep going

3. Thread Author
$$\displaystyle 16\left[ -\frac{\ln\left({\left| x-2 \right|}\right)}{4} +\frac{\ln\left({\left| x+2 \right|}\right)}{4} \right] +\frac{{x}^{3}}{3} +4x$$
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. Thread Author
What was inside the brackets got turned around making the leading term positive

6. Originally Posted by Prove It
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. Originally Posted by greg1313
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. Thread Author
$\textsf{Evaluate using decomposition}$
\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...

9. Originally Posted by 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...
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*}

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