How do I find the integral of x^3/(x^2+4)?

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  • Thread starter karush
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In summary, the expression "W8.5.7 int x^3/(x^2 +4)" is a mathematical function used to calculate the integral of x^3 divided by the sum of x^2 and 4. The integral is calculated using the power rule and the sum rule, resulting in a function of x. The expression can be evaluated for any real number value of x and can be simplified through algebraic manipulation or using a graphing calculator. The number 8.5.7 is used to identify and differentiate the integral from other similar expressions in a textbook or mathematical resource.
  • #1
karush
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W8.5.7
Evaluate
$$\displaystyle \int\frac{{x}^{3}}{{x}^{2 }+4}\ dx $$
The book answer was
$$\displaystyle\frac{{x}^{2}}{2 }
-2\ln\left({4+{x}^{2 }}\right)+C$$
So
$$\displaystyle u=x^2 +4\ \ \ \ du=2x\ \ dx \ \ x=\sqrt{u-4} $$
Then
$$\displaystyle \int\frac{{x}^{2}}{{x}^{2 }+4}x\ dx
\implies \frac{1}{2} \int \frac{u-4}{u}\ \ du$$
I continued but didn't get the answer
 
Last edited:
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  • #2
If you use:

\(\displaystyle u=x^2+4\)

then we have:

\(\displaystyle x=\sqrt{u-4}\implies dx=\frac{1}{2\sqrt{u-4}}\,du\)

And so the integral becomes:

\(\displaystyle I=\frac{(u-4)^{\frac{3}{2}}}{u}\cdot\frac{1}{2\sqrt{u-4}}\,du=\frac{1}{2}\int \frac{u-4}{u}\,du=\frac{1}{2}\int 1-\frac{4}{u}\,du\)

When you integrate, combine the resulting constant with the constant of integration and then back-substitute for $u$.
 
  • #3
$$ \frac{1}{2 }\int 1\ du - 2 \int\frac{1}{u} \ du
\implies\frac{u}{2}-2 \ln\left({u}\right)+C$$
Plug in is close but not it
 
Last edited:
  • #4
karush said:
$$ \frac{1}{2 }\int 1\ dx - 2 \int\frac{1}{u} \ du
\implies\frac{u}{2}-2 \ln\left({u}\right)+C$$
Plug in is close but not it

When you back-substitute for $u$, what do you get?
 
  • #5
$$\frac{x^2 +4}{2} +2\ln\left({x^2 +4}\right)+C $$
The 4 in the leading term shouldn't be there
 
  • #6
karush said:
$$\frac{x^2 +4}{2} +2\ln\left({x^2 +4}\right)+C $$
The 4 in the leading term shouldn't be there

Then separate it, and move it like so:

\(\displaystyle \frac{x^2}{2} +2\ln\left({x^2 +4}\right)+C+2\)

Now, an arbitrary constant $C$ with $2$ added to it is still just an arbitrary constant, so you may write:

\(\displaystyle \frac{x^2}{2} +2\ln\left({x^2 +4}\right)+C\)

When doing indefinite integrals, if your anti-derivative differs from the given answer by only a constant, then you know the two results are equivalent. :)
 
  • #7
Well sure helpful to know that.. 😍
 
  • #8
karush said:
W8.5.7
Evaluate
$$\displaystyle \int\frac{{x}^{3}}{{x}^{2 }+4}\ dx $$
The book answer was
$$\displaystyle\frac{{x}^{2}}{2 }
-2\ln\left({4+{x}^{2 }}\right)+C$$
So
$$\displaystyle u=x^2 +4\ \ \ \ du=2x\ \ dx \ \ x=\sqrt{u-4} $$
Then
$$\displaystyle \int\frac{{x}^{2}}{{x}^{2 }+4}x\ dx
\implies \frac{1}{2} \int \frac{u-4}{u}\ \ du$$
I continued but didn't get the answer

Whenever you have a rational function of polynomials with the numerator of higher degree than the denominator, a good simplification is to long divide.

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

Both of these resulting integrals should be VERY easy to solve...
 
  • #9
$\displaystyle\frac{{x}^{2}}{2 }
-2\ln\left({4+{x}^{2 }}\right)+C$
 

FAQ: How do I find the integral of x^3/(x^2+4)?

What is the purpose of the expression "W8.5.7 int x^3/(x^2 +4)"?

The expression represents a mathematical function, which is used to calculate the integral of x^3 divided by the sum of x^2 and 4.

How is the integral calculated in "W8.5.7 int x^3/(x^2 +4)"?

The integral is calculated using the rules of integration, specifically the power rule and the sum rule. The result will be a function of x.

Is there a specific range of values for x in "W8.5.7 int x^3/(x^2 +4)"?

No, the expression can be evaluated for any real number value of x.

Can the expression "W8.5.7 int x^3/(x^2 +4)" be simplified?

Yes, it can be simplified by using algebraic manipulation or by using a graphing calculator to approximate the integral.

What is the significance of the number 8.5.7 in "W8.5.7 int x^3/(x^2 +4)"?

The number 8.5.7 represents the specific problem or exercise in a textbook or mathematical resource. It is used to identify and differentiate the integral from other similar expressions.

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