What is the Integral Expansion for UHWO 242?

In summary, the original integral can be solved by using partial fractions to split it into three integrals. These integrals can be solved using different substitution methods, resulting in the final answer of $\displaystyle \frac{3}{2}\ln{(x^2+1)} - 3\arctan{(x)} + \sqrt{2}\arctan{\left(\frac{x}{\sqrt{2}}\right)} + C$.
  • #1
karush
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$$\displaystyle
I=\int \frac{3{x}^{2}-{x}^{2}+6x-4}{({x}^{2}+1)({x}^{2}+2)} \\
=\frac{2}{3}\ln\left({x}^{2}+1\right)-3\arctan\left({x}\right)
+\sqrt{2}\arctan\left({x/\sqrt{2}}\right)+C$$

Was going to take this a step at a time
so by observation this looks like an expasion
The way the answer looks
 
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  • #2
\(\displaystyle \dfrac{3x^3-x^2+6x-4}{(x^2+1)(x^2+2)}=\dfrac{3x-3}{x^2+1}+\dfrac{2}{x^2+2}\)
 
  • #3
I got but think its the same

$$\frac{2}{\left(x^{2}+2\right)}
+\frac{3x}{\left(x^{2}+1\right)}
-\frac{3}{\left(x^{2}+1\right)}$$
 
  • #4
It's the same.
 
  • #5
karush said:
$$\displaystyle
I=\int \frac{3{x}^{2}-{x}^{2}+6x-4}{({x}^{2}+1)({x}^{2}+2)} \\
=\frac{2}{3}\ln\left({x}^{2}+1\right)-3\arctan\left({x}\right)
+\sqrt{2}\arctan\left({x/\sqrt{2}}\right)+C$$

Was going to take this a step at a time
so by observation this looks like an expasion
The way the answer looks

I'm assuming it's actually $\displaystyle \begin{align*} \int{ \frac{3\,x^3 - x^2 + 6\,x - 4}{\left( x^2 + 1 \right) \left( x^2 + 2 \right) } \,\mathrm{d}x} \end{align*}$, anyway applying Partial Fractions...

$\displaystyle \begin{align*} \frac{A\,x + B}{x^2 + 1} + \frac{C\,x + D}{x^2 + 2} &\equiv \frac{3\,x^3 - x^2 + 6\,x - 4}{\left( x^2 + 1 \right) \left( x^2 + 2 \right) } \\ \frac{ \left( A\,x + B \right) \left( x^2 + 2 \right) + \left( C\,x + D \right) \left( x^2 + 1 \right) }{ \left( x^2 + 1 \right) \left( x^2 + 2 \right) } &\equiv \frac{3\,x^3 - x^2 + 6\,x - 4}{ \left( x^2 + 1 \right) \left( x^2 + 2 \right) } \\ \left( A\,x + B \right) \left( x^2 + 2 \right) + \left( C\,x + D \right) \left( x^2 + 1 \right) &\equiv 3\,x^3 - x^2 + 6\,x - 4 \end{align*}$

Let $\displaystyle \begin{align*} x = \mathrm{i} \end{align*}$ to find $\displaystyle \begin{align*} A\,\mathrm{i} + B = 3\,\mathrm{i} - 3 \implies A = 3 \textrm{ and } B = -3 \end{align*}$.

Let $\displaystyle \begin{align*} x = \sqrt{2}\,\mathrm{i} \end{align*}$ to find $\displaystyle \begin{align*} -\sqrt{2}\,\mathrm{i}\,C - D = 0\,\mathrm{i} - 2 \implies C = 0 \textrm{ and } D = 2 \end{align*}$. So

$\displaystyle \begin{align*} \int{ \frac{3\,x^3 - x^2 + 6\,x - 4}{\left( x^2 + 1 \right) \left( x^2 + 2 \right) } \,\mathrm{d}x } &= \int{ \left( \frac{3\,x - 3}{x^2 + 1} + \frac{2}{x^2 + 2} \right) \,\mathrm{d}x } \\ &= \frac{3}{2} \int{ \frac{2\,x - 2}{x^2 + 1} \,\mathrm{d}x } + 2 \int{ \frac{1}{x^2 + 2} \,\mathrm{d}x } \\ &= \frac{3}{2} \int{ \frac{2\,x}{x^2 + 1} \,\mathrm{d}x } - 3\int{ \frac{1}{x^2 + 1}\,\mathrm{d}x } + 2\int{ \frac{1}{x^2 + 2} \,\mathrm{d}x } \end{align*}$

The first integral can be solved with the substitution $\displaystyle \begin{align*} u = x^2 + 1 \implies \mathrm{d}u = 2\,x\,\mathrm{d}x \end{align*}$, the second can be solved with the substitution $\displaystyle \begin{align*} x = \tan{ \left( \theta \right) } \implies \mathrm{d}x = \sec^2{ \left( \theta \right) } \,\mathrm{d}\theta \end{align*}$, and the third with the substitution $\displaystyle \begin{align*} x = \sqrt{2}\tan{ \left( t \right) } \implies \mathrm{d}x = \sqrt{2}\sec^2{ \left( t \right) }\,\mathrm{d}t \end{align*}$.
 
  • #6
$$\displaystyle
I= \int{ \frac{3\,x^3 - x^2 + 6\,x - 4}{\left( x^2 + 1 \right) \left( x^2 + 2 \right) } \,dx } \\
= \frac{3}{2} \int{ \frac{2\,x}{x^2 + 1} \,dx }
- 3\int{ \frac{1}{x^2 + 1}\,dx }
+ 2\int{ \frac{1}{x^2 + 2} \,dx } $$
solving (1)
$$\displaystyle\frac{3}{2} \int{ \frac{2\,x}{x^2 + 1} \,dx } =
\frac{3\ln\left({{x}^{2}+1}\right)}{2}$$
solving (2)
$$- 3\int{ \frac{1}{x^2 + 1}\,dx } \\
\displaystyle \begin{align*} x = \tan{ \left( u \right) }
\implies dx = \sec^2{ \left(u \right) } \,du
\end{align*} \\
-3\int\frac{1}{\tan^2\left({u}\right)+1}\sec^2{ \left(u \right) } \,du
$$
doesn't this cancel out
 
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FAQ: What is the Integral Expansion for UHWO 242?

What is UHWO 242 Integral expansion?

UHWO 242 Integral expansion is a mathematical concept in the field of integral calculus. It involves breaking down a complex integral into simpler integrals, often using methods such as partial fractions or integration by parts.

What are some real-life applications of UHWO 242 Integral expansion?

UHWO 242 Integral expansion is used in various fields such as physics, engineering, and economics. Some examples of real-life applications include calculating the area under a curve, computing work done by a varying force, and estimating the value of a definite integral.

How is UHWO 242 Integral expansion different from other methods of integration?

UHWO 242 Integral expansion is a specific technique used to solve certain types of integrals. It is different from other methods such as substitution or trigonometric substitution, which are used to solve different types of integrals.

Can UHWO 242 Integral expansion be used to solve all integrals?

No, UHWO 242 Integral expansion is not applicable to all integrals. It is most commonly used for integrals that involve rational functions, but it may not work for integrals with more complex functions or expressions.

Is UHWO 242 Integral expansion difficult to learn?

UHWO 242 Integral expansion can be challenging to understand at first, but with practice and guidance, it can be mastered. It is important to have a strong foundation in basic integration techniques before attempting to learn UHWO 242 Integral expansion.

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