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

[karush]

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

 

[MarkFL]

If you use:

$$u=x^2+4$$

then we have:

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

And so the integral becomes:

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

When you integrate, combine the resulting constant with the constant of integration and then back-substitute for $u$.

 

[karush]

$$ \frac{1}{2 }\int 1\ du - 2 \int\frac{1}{u} \ du
\implies\frac{u}{2}-2 \ln\left({u}\right)+C$$
Plug in is close but not it

 

[MarkFL]

$$ \frac{1}{2 }\int 1\ dx - 2 \int\frac{1}{u} \ du
\implies\frac{u}{2}-2 \ln\left({u}\right)+C$$
Plug in is close but not it

When you back-substitute for $u$, what do you get?

 

[karush]

$$\frac{x^2 +4}{2} +2\ln\left({x^2 +4}\right)+C $$
The 4 in the leading term shouldn't be there

 

[MarkFL]

$$\frac{x^2 +4}{2} +2\ln\left({x^2 +4}\right)+C $$
The 4 in the leading term shouldn't be there

Then separate it, and move it like so:

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

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

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

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. :)

 

[karush]

Well sure helpful to know that..

 

[Prove It]

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...

 

[karush]

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