MHB How Do You Express and Evaluate This Integral Using Partial Fractions?

[karush]

$\tiny{242 .10.09.8}\\$
$\textsf{Express the integrand as a sum of partial fractions and evaluate integral}$
\begin{align*}\displaystyle
I&=\int f \, dx = \int\frac{\sqrt{16+5x}}{x} \, dx
\end{align*}
\begin{align*}\displaystyle
f&=\frac{\sqrt{16+5x}}{x}
=\frac{\sqrt{16+5x}}{x}\cdot\frac{\sqrt{16+5x}}{\sqrt{16+5x}}\\
&=\frac{16+5x}{x\sqrt{16+5x}}=\frac{16}{x}+\frac{5x}{\sqrt{16+5x}}
\end{align*}
$\textsf{ok, I did something ??}\\$
$\textsf{Answer from}$ $ \textbf{W|A}$
$$\displaystyle I= 2\sqrt{5x+16}+4\log(4-\sqrt{5x+16})-4\log(\sqrt{5x-16}+4)+C$$

 

[MarkFL]

I think what I would do is use the substitution:

$$16+5x=u^2\implies dx=\frac{2}{5}u\,du$$

And now the integral becomes:

$$I=2\int \frac{u^2}{u^2-16}\,du=2\int 1+\frac{2}{u-4}-\frac{2}{u+4}\,du$$

 

[karush]

$\tiny{7.6.16}$
$\textit{Then}\\$
$$\displaystyle I=2\left[\int 1 \,du +2\int\frac{1}{u-4}\,du
-2\int\frac{1}{u+4}\,du \right]+C$$
$\textit{Then Integrate}\\$
$$I=2u+4\ln(u-4)-4\ln(u+4)+C$$
$\textit{Then substute back in $u=\sqrt{16+5x}$}$
$$\displaystyle I= 2\sqrt{5x+16}+4\log(\sqrt{5x+16}-4)-4\log(\sqrt{5x-16}+4)+C$$

 

[MarkFL]

$\tiny{7.6.16}$
$\textit{Then}\\$
$$\displaystyle I=2\left[\int 1 \,du +2\int\frac{1}{u-4}\,du
-2\int\frac{1}{u+4}\,du \right]+C$$
$\textit{Then Integrate}\\$
$$I=2u+4\ln(u-4)-4\ln(u+4)+C$$
$\textit{Then substute back in $u=\sqrt{16+5x}$}$
$$\displaystyle I= 2\sqrt{5x+16}+4\log(\sqrt{5x+16}-4)-4\log(\sqrt{5x-16}+4)+C$$

Can you explain what minor error in notation used by both you and W|A has led to different answers?

 

[Prove It]

$\tiny{242 .10.09.8}\\$
$\textsf{Express the integrand as a sum of partial fractions and evaluate integral}$
\begin{align*}\displaystyle
I&=\int f \, dx = \int\frac{\sqrt{16+5x}}{x} \, dx
\end{align*}
\begin{align*}\displaystyle
f&=\frac{\sqrt{16+5x}}{x}
=\frac{\sqrt{16+5x}}{x}\cdot\frac{\sqrt{16+5x}}{\sqrt{16+5x}}\\
&=\frac{16+5x}{x\sqrt{16+5x}}=\frac{16}{x}+\frac{5x}{\sqrt{16+5x}}
\end{align*}
$\textsf{ok, I did something ??}\\$
$\textsf{Answer from}$ $ \textbf{W|A}$
$$\displaystyle I= 2\sqrt{5x+16}+4\log(4-\sqrt{5x+16})-4\log(\sqrt{5x-16}+4)+C$$

If you are going to try to use algebraic manipulation to perform an integral, you MUST do the algebra correctly!

$\displaystyle \begin{align*} \int{ \frac{16 + 5\,x}{x\,\sqrt{16 + 5\,x}}\,\mathrm{d}x } &= \int{ \left( \frac{16}{x\,\sqrt{16 + 5\,x}} + \frac{5\,x}{x\,\sqrt{16 + 5\,x}} \right) \,\mathrm{d}x } \\ &= \int{ \left( \frac{16}{x\,\sqrt{16 + 5\,x}} + \frac{5}{\sqrt{16 + 5\,x}} \right) \,\mathrm{d}x } \\ &= \int{ \frac{16}{x\,\sqrt{16 + 5\,x}} \,\mathrm{d}x } + \int{ \frac{5}{\sqrt{16 + 5\,x}} \,\mathrm{d}x } \\ &= \frac{16}{5} \int{ \frac{5}{x\,\sqrt{16 - 5\,x}} \,\mathrm{d}x } + \int{ \frac{5}{\sqrt{16 + 5\,x}}\,\mathrm{d}x } \end{align*}$

Now let $\displaystyle \begin{align*} u = 16 + 5\,x \implies \mathrm{d}u = 5\,\mathrm{d}x \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \frac{16}{5}\int{ \frac{5}{x\,\sqrt{16 + 5\,x}}\,\mathrm{d}x } + \int{ \frac{5}{\sqrt{16 + 5\,x}} \,\mathrm{d}x } &= \frac{16}{5}\int{ \frac{1}{\left( \frac{u - 16}{5} \right) \,\sqrt{u}}\,\mathrm{d}u } + \int{
\frac{1}{\sqrt{u}} \,\mathrm{d}u } \\ &= 16 \int{ \frac{1}{\left[ \left( \sqrt{u} \right) ^2 - 16 \right] \,\sqrt{u}} \,\mathrm{d}u } + \int{u^{-\frac{1}{2}}\,\mathrm{d}u} \\ &= 32 \int{ \left[ \frac{1}{\left( \sqrt{u} \right) ^2 - 16 } \right] \,\frac{1}{2\,\sqrt{u}} \,\mathrm{d}u } + \frac{u^{\frac{1}{2}}}{\frac{1}{2}} + C_1 \end{align*}$

So now let $\displaystyle \begin{align*} v = \sqrt{u} \implies \mathrm{d}v = \frac{1}{2\,\sqrt{u}}\,\mathrm{d}u \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} 32\int{ \left[ \frac{1}{ \left( \sqrt{u} \right) ^2 - 16 } \right] \,\frac{1}{2\,\sqrt{u}}\,\mathrm{d}u } + \frac{u^{\frac{1}{2}}}{\frac{1}{2}} + C_1 &= 32 \int{ \frac{1}{v^2 - 16} \,\mathrm{d}v } + 2\,\sqrt{u} + C_1 \\ &= 32 \int{ \frac{1}{ \left( v - 4 \right) \left( v + 4 \right) } \,\mathrm{d}v } + 2\,\sqrt{16 + 5\,x} + C_1 \end{align*}$

and now it's in a form you can use Partial Fractions for...

 

[karush]

Can you explain what minor error in notation used by both you and W|A has led to different answers?

I presume it was the 3rd term should be

$$-4\log(\sqrt{5x+16}+4)$$

 

[MarkFL]

I presume it was the 3rd term should be

$$-4\log(\sqrt{5x+16}+4)$$

No, it's the second term, which W|A gives as:

$$4\log(4-\sqrt{5x+16})$$

But, you gave as:

$$4\log(\sqrt{5x-16}-4)$$

You should confirm that these two different terms do not differ by a constant. The issue here is that what should be used is:

$$\int \frac{du}{u}=\ln|u|+C$$