# What is int 9/(x^2sqrt(4x+1))dx?

## I found this integral question I asked in Yahoo! Answers over 4 years ago. It was answered, but I think we can do better!

Dec 25, 2016

I got:

$18 \ln | \frac{\sqrt{4 x + 1} + 1}{\sqrt{4 x + 1} - 1} | - \frac{9 \sqrt{4 x + 1}}{x} + C$

and to prove that it worked, Wolfram Alpha shows that the derivative of this is the original integrand for $x > 0$.

Let's take a crack at this. I can expect some form of substitution, and probably partial fractions.

Let:

$u = \sqrt{4 x + 1}$

$\mathrm{du} = \frac{2}{\sqrt{4 x + 1}} \mathrm{dx}$
$\implies \mathrm{dx} = \frac{u}{2} \mathrm{du}$

${x}^{2} = {\left({u}^{2} - 1\right)}^{2} / 16$

This gives:

$\implies 9 \int \frac{1}{{x}^{2} \sqrt{4 x + 1}} \mathrm{dx}$

$= 9 \int \frac{16}{{\left({u}^{2} - 1\right)}^{2} \cancel{u}} \frac{\cancel{u}}{2} \mathrm{du}$

$= 72 \int \frac{1}{{\left({u}^{2} - 1\right)}^{2}} \mathrm{du}$

Yep, partial fractions. With this, we have:

$\int \frac{1}{{u}^{2} - 1} ^ 2$

$= \int \frac{A}{u + 1} + \frac{B}{u + 1} ^ 2 + \frac{C}{u - 1} + \frac{D}{u - 1} ^ 2 \mathrm{du}$

Getting common denominators allows us to cancel out the denominators and focus on the numerator:

$\implies A \left(u + 1\right) {\left(u - 1\right)}^{2} + B {\left(u - 1\right)}^{2} + C \left(u - 1\right) {\left(u + 1\right)}^{2} + D {\left(u + 1\right)}^{2}$

$= \left(A u + A\right) \left({u}^{2} - 2 u + 1\right) + B {u}^{2} - 2 B u + B + \left(C u - C\right) \left({u}^{2} + 2 u + 1\right) + D {u}^{2} + 2 D u + D$

$= \textcolor{\mathrm{da} r k b l u e}{A {u}^{3}} - \textcolor{\mathmr{and} a n \ge}{2 A {u}^{2}} + \textcolor{c y a n}{A u} + \textcolor{\mathmr{and} a n \ge}{A {u}^{2}} - \textcolor{c y a n}{2 A u} + \textcolor{red}{A} + \textcolor{\mathmr{and} a n \ge}{B {u}^{2}} - \textcolor{c y a n}{2 B u} + \textcolor{red}{B} + \textcolor{\mathrm{da} r k b l u e}{C {u}^{3}} + \textcolor{\mathmr{and} a n \ge}{2 C {u}^{2}} + \textcolor{c y a n}{C u} - \textcolor{\mathmr{and} a n \ge}{C {u}^{2}} - \textcolor{c y a n}{2 C u} - \textcolor{red}{C} + \textcolor{\mathmr{and} a n \ge}{D {u}^{2}} + \textcolor{c y a n}{2 D u} + \textcolor{red}{D}$

Collect like terms (making sure terms are grouped by addition!) to get:

$= A {u}^{3} + C {u}^{3} {\textcolor{red}{\cancel{\textcolor{b l a c k}{- 2 A {u}^{2} + A {u}^{2}}}}}^{- A {u}^{2}} + B {u}^{2} + {\textcolor{red}{\cancel{\textcolor{b l a c k}{2 C {u}^{2} - C {u}^{2}}}}}^{C {u}^{2}} + D {u}^{2} + {\textcolor{red}{\cancel{\textcolor{b l a c k}{A u - 2 A u}}}}^{- A u} - 2 B u + {\textcolor{red}{\cancel{\textcolor{b l a c k}{C u - 2 C u}}}}^{- C u} + 2 D u + A + B - C + D$

$\implies \left(A + C\right) {u}^{3} + \left(- A + B + C + D\right) {u}^{2} + \left(- A - 2 B - C + 2 D\right) u + \left(A + B - C + D\right) = 0 {u}^{3} + 0 {u}^{2} + 0 u + 1$

This means we have a system of equations:

$A + 0 B + C + 0 D = 0$
$- A + B + C + D = 0$
$- A - 2 B - C + 2 D = 0$
$A + B - C + D = 1$

Or, the matrix:

$\left[\begin{matrix}1 & 0 & 1 & 0 & | & 0 \\ - 1 & 1 & 1 & 1 & | & 0 \\ - 1 & - 2 & - 1 & 2 & | & 0 \\ 1 & 1 & - 1 & 1 & | & 1\end{matrix}\right]$

where the first four columns imply the variable $A$, $B$, $C$, or $D$, respectively, and the last column is the righthand side of the original system of equations.

We can use elementary row operations to simplify this matrix into one that gives a more obvious answer.

Let $c {R}_{i} + {R}_{j}$ denote multiplying row $i$ by a constant $c$, adding the result to row $j$, and storing it in row $j$. Then:

$\stackrel{{R}_{1} + {R}_{2} \text{ }}{\to} \left[\begin{matrix}1 & 0 & 1 & 0 & | & 0 \\ 0 & 1 & 2 & 1 & | & 0 \\ - 1 & - 2 & - 1 & 2 & | & 0 \\ 1 & 1 & - 1 & 1 & | & 1\end{matrix}\right]$

$\stackrel{{R}_{1} + {R}_{3} \text{ }}{\to} \left[\begin{matrix}1 & 0 & 1 & 0 & | & 0 \\ 0 & 1 & 2 & 1 & | & 0 \\ 0 & - 2 & 0 & 2 & | & 0 \\ 1 & 1 & - 1 & 1 & | & 1\end{matrix}\right]$

$\stackrel{- {R}_{1} + {R}_{4} \text{ }}{\to} \left[\begin{matrix}1 & 0 & 1 & 0 & | & 0 \\ 0 & 1 & 2 & 1 & | & 0 \\ 0 & - 2 & 0 & 2 & | & 0 \\ 0 & 1 & - 2 & 1 & | & 1\end{matrix}\right]$

$\stackrel{- {R}_{2} + {R}_{4} \text{ }}{\to} \left[\begin{matrix}1 & 0 & 1 & 0 & | & 0 \\ 0 & 1 & 2 & 1 & | & 0 \\ 0 & - 2 & 0 & 2 & | & 0 \\ 0 & 0 & - 4 & 0 & | & 1\end{matrix}\right]$

This is far enough to find that:

$- 4 C = 1 \implies C = - 1 \text{/} 4$

$- 2 B + 2 D = 0 \implies B = D$

$B + 2 C + D = 0 \implies C = - D = - B$

$A = - C$

Therefore, our solutions are:

$\textcolor{g r e e n}{A = B = D = 1 \text{/} 4}$
$\textcolor{g r e e n}{C = - 1 \text{/} 4}$

That means:

$\int \frac{9}{{x}^{2} \sqrt{4 x + 1}} \mathrm{dx} = 72 \int \frac{1}{{u}^{2} - 1} ^ 2 \mathrm{du}$

$= \frac{72}{4} \int \frac{1}{u + 1} + \frac{1}{u + 1} ^ 2 - \frac{1}{u - 1} + \frac{1}{u - 1} ^ 2 \mathrm{du}$

$= 18 \left[\ln | u + 1 | - \frac{1}{u + 1} - \ln | u - 1 | - \frac{1}{u - 1}\right]$

$= 18 \left[\ln | \frac{u + 1}{u - 1} | - \frac{u - 1}{{u}^{2} - 1} - \frac{u + 1}{{u}^{2} - 1}\right]$

$= 18 \left[\ln | \frac{u + 1}{u - 1} | - \left(\frac{u - 1}{{u}^{2} - 1} + \frac{u + 1}{{u}^{2} - 1}\right)\right]$

$= 18 \left[\ln | \frac{u + 1}{u - 1} | - \frac{2 u}{{u}^{2} - 1}\right]$

$= 18 \ln | \frac{u + 1}{u - 1} | - \frac{36 u}{{u}^{2} - 1}$

Finally, substitute back in $u = \sqrt{4 x + 1}$ to get:

$\implies \textcolor{b l u e}{\int \frac{9}{{x}^{2} \sqrt{4 x + 1}} \mathrm{dx}}$

$= \textcolor{b l u e}{18 \ln | \frac{\sqrt{4 x + 1} + 1}{\sqrt{4 x + 1} - 1} | - \frac{9 \sqrt{4 x + 1}}{x} + C}$

Yep, still hard, even today!

Jan 14, 2017

If we hate partial fractions as I do, we can follow the method @truong-son-n used with the substitution $t = \sqrt{4 x + 1}$ up to the point:

$\int \frac{9}{{x}^{2} \sqrt{4 x + 1}} \mathrm{dx} = 72 \int \frac{1}{{t}^{2} - 1} ^ 2 \mathrm{dt}$

Now apply the substitution $t = \sec \theta$. Thus $\mathrm{dt} = \sec \theta \tan \theta d \theta$.

$= 72 \int \frac{\sec \theta \tan \theta}{{\sec}^{2} \theta - 1} ^ 2 d \theta$

Since ${\sec}^{2} \theta - 1 = {\tan}^{2} \theta$:

$= 72 \int \frac{\sec \theta \tan \theta}{\tan} ^ 4 \theta d \theta = 72 \int \sec \frac{\theta}{\tan} ^ 3 \theta = 72 \int {\cos}^{2} \frac{\theta}{\sin} ^ 3 \theta = 72 \int {\cot}^{2} \theta \csc \theta d \theta$

From ${\cot}^{2} \theta = {\csc}^{2} \theta - 1$:

$= 72 \int {\csc}^{3} \theta d \theta - 72 \int \csc \theta d \theta$

Letting $I = \int {\csc}^{3} \theta d \theta$, we will perform integration by parts letting

$\left\{\begin{matrix}u = \csc \theta \text{ "=>" "du=-cscthetacotthetad theta \\ dv=csc^2thetad theta" "=>" } v = - \cot \theta\end{matrix}\right.$

So

$I = - \csc \theta \cot \theta - \int \csc \theta {\cot}^{2} \theta d \theta$

Again using ${\cot}^{2} \theta = {\csc}^{2} \theta - 1$:

$I = - \csc \theta \cot \theta - \int {\csc}^{3} \theta d \theta + \int \csc \theta d \theta$

Using $I = {\csc}^{3} \theta$, that is, noticing the original integral is back into the mix, and also using the fairly well-known integral $\int \csc \theta d \theta = - \ln \left\mid \cot \theta + \csc \theta \right\mid$ we see that

$2 I = - \csc \theta \cot \theta - \ln \left\mid \cot \theta + \csc \theta \right\mid$

$\int {\csc}^{3} \theta d \theta = - \frac{1}{2} \csc \theta \cot \theta - \frac{1}{2} \ln \left\mid \cot \theta + \csc \theta \right\mid$

Which we will plug back into our original integral:

$= 72 \left(- \frac{1}{2} \csc \theta \cot \theta - \frac{1}{2} \ln \left\mid \cot \theta + \csc \theta \right\mid\right) - 72 \int \csc \theta d \theta$

$= - 36 \csc \theta \cot \theta - 36 \ln \left\mid \cot \theta + \csc \theta \right\mid + 72 \ln \left\mid \cot \theta + \csc \theta \right\mid$

$= - 36 \csc \theta \cot \theta + 36 \ln \left\mid \cot \theta + \csc \theta \right\mid$

Our substitution was $t = \sec \theta$. This is a triangle where $t$ is the hypotenuse, $1$ is the adjacent side, and $\sqrt{{t}^{2} - 1}$ is the opposite side.

Thus $\csc \theta = \frac{t}{\sqrt{{t}^{2} - 1}}$ and $\cot \theta = \frac{1}{\sqrt{{t}^{2} - 1}}$.

Plugging these in gives:

$= - \frac{36 t}{{t}^{2} - 1} + 36 \ln \left\mid \frac{t + 1}{\sqrt{{t}^{2} - 1}} \right\mid$

With $t = \sqrt{4 x + 1}$:

$= - \frac{36 \sqrt{4 x + 1}}{\left(4 x + 1\right) - 1} + 36 \ln \left\mid \frac{\sqrt{4 x + 1} + 1}{\sqrt{\left(4 x + 1\right) - 1}} \right\mid$

$= - \frac{9 \sqrt{4 x + 1}}{x} + 36 \ln \left\mid \frac{\sqrt{4 x + 1} + 1}{2 \sqrt{x}} \right\mid + C$

Which, when graphically confirmed, is the same answer that Truong-Son found.