# Solving the integrand $\frac{x^3}{\sqrt{x^2+10x+16}}$

I just wanted to make sure that what I did to integrate $$\frac{x^3}{\sqrt{x^2+10x+16}}$$ is correct.

I assumed that it is classified as a trigonometric substitution problem. And so, what I first did is to apply "completing the square":

$$\int \frac{x^3 dx}{\sqrt{x^2+10x+16}} = \int \frac{x^3 dx}{\sqrt{x^2+10x+25+16-25}} = \int \frac{x^3 dx}{\sqrt{(x+5)^2-9}}$$

After that, I assigned values into some variables:

let $$a = 3$$

$$x + 5 = 3 \sec \Theta \rightarrow \sec \Theta = \frac{x+5}{3}$$

$$dx = 3 \sec \Theta \tan \Theta\, d \Theta$$

Next, I have substituted the value of (x+5) to $$\sqrt{(x+5)^2-9}$$ which leads to $$3 \tan \Theta$$. And, $$\tan \Theta$$ is equal to $$\frac{\sqrt{x^2+10x+16}}{3}$$.

Afterwards, I have replaced all of the variables to the values that I assigned to them:

$$\int \frac{x^3}{\sqrt{x^2+10x+16}} \rightarrow \int \frac{(3 \sec \Theta)^3 (3 \sec \Theta \tan \Theta)\, d\Theta}{3 \tan \Theta} \rightarrow \int (3 \sec\Theta - 5)^3 \sec\Theta\, d \Theta$$

Expanding the trinomial, distributing $$\sec \theta$$ to each term and applying the constant theorem will result to:

$$27\int \sec^4\Theta d\Theta - 135\int sec^3\Theta\, d\Theta + 225 \int sec^2\Theta d\Theta - 125 \int sec \Theta d \Theta$$

Now, by making $$\sec^2 \Theta$$ into $$(1 + \tan^2 \Theta)$$ and applying u-substitution: $$27 \int \sec^4 \Theta \,d\Theta \rightarrow 27 \tan \Theta + 9 \tan^3 \Theta + C$$

By applying integration by parts, $$\int \sec^3 \Theta\, d\Theta$$ would become $$\frac{\sec\Theta tan\Theta + \ln\left | \sec\Theta + \tan\Theta \right |}{2}$$.

Finally, $$\int \sec^2 \Theta\, d\Theta$$ would simply become $$\tan \Theta$$ and $$\int \sec \Theta$$ would be $$ln\left | \sec\Theta + \tan\Theta \right |$$ .

Since $$\sec \Theta$$ is equal to $$\frac{x+5}{3}$$ and $$\tan\Theta$$ is equal to $$\frac{\sqrt{x^2+10x+16}}{3}$$, the whole integral would be (I had added all like terms before this):

$$9\left [ \frac{\sqrt{x^2+10x+16}}{3} \right ]^3 + 252\left ( \frac{\sqrt{x^2+10x+16}}{3}\right ) -\frac{135}{2} \left ( \frac{x+5}{3}\right ) \left ( \frac{\sqrt{x^2+10x+16}}{3}\right ) - \frac{385}{2} ln \left | \frac{x+5}{3} + \frac{\sqrt{x^2+10x+16}}{3}\right | + C$$

Lastly, I simplified the integral:

$$\frac{1}{3} \left ( x^2 + 10x+16 \right )^\frac{3}{2} + \frac{(93-15x)\sqrt{x^2+10x+16}}{2} - \frac{385}{2} ln \left | \frac{x+5+\sqrt{x^2+10x+16}}{3}\right | + C$$

Have I integrated the integrand appropriately? Thanks in advanced.

• No. How did you eliminate the square root in the first step? By the way, you should get: $$\frac{1}{6} \sqrt{x^2+10 x+16} \left(2 x^2-25 x+311\right)-\frac{385}{2} \log \left(\sqrt{x^2+10 x+16}+x+5\right))$$ Commented Dec 24, 2019 at 1:59
• It's a typo. I edited it. Commented Dec 24, 2019 at 2:20
• @DavidG.Stork, he is getting this only, if you combine first two terms. Commented Dec 24, 2019 at 2:38

It is correct. Here is another method without trigonometric substitution: $$\int\frac{x^3}{\sqrt{x^2+10x+16}}dx=\int\frac{x^3+10x^2+16x-10x^2-100x-160+84x+160}{\sqrt{x^2+10x+16}}dx$$ $$= \int\frac{(x-10)(x^2+10x+16)+84x+160}{\sqrt{x^2+10x+16}}dx$$ $$= \int(x-10)\sqrt{x^2+10x+16}\ dx+\int\frac{84x+420}{\sqrt{x^2+10x+16}}dx-260\int\frac{1}{\sqrt{x^2+10x+16}}dx$$ The last integral can be directly solved using formula, second last one will follow if you substitute $$t=x^2+10x+16$$, remains to solve the first integral. $$\int(x-10)\sqrt{x^2+10x+16}\ dx = \frac{1}{2}\int(2x+10)\sqrt{x^2+10x+16}\ dx-15\int\sqrt{x^2+10x+16}\ dx$$ Last integral follows directly from formula and second last is solved by substituting $$x^2+10x+16$$.

Note that this is a general method. In case there is any polynomial in numerator and any quadratic (with or without square root) in denominator, divide the numerator by denominator to get a linear in numerator, then reduce the numerator to a constant times the derivative of quadratic plus another constant, which can be solved easily using substitution and existing formulas.

To evaluate the integral systematically, apply the reduction formula \begin{align} I_n=& \int\frac{x^n}{\sqrt{x^2+10x+16}}dx\\ =&\ \frac{x^{n-1}}n \sqrt{x^2+10x+16} -\frac{5(2n-1)}n I_{n-1}-\frac{16(n-1)}n I_{n-2} \end{align} to simplify \begin{align} \int\frac{x^3}{\sqrt{x^2+10x+16}}dx = \left(\frac{x^{2}}3-\frac{25x}6+\frac{311}6 \right)\sqrt{x^2+10x+16} -\frac{285}2 I_0 \end{align} where $$I_0=\int\frac{1}{\sqrt{x^2+10x+16}}dx =\coth^{-1}\frac{x+5}{\sqrt{x^2+10x+16}}$$

I might start differently, completing the square:

$$\frac{x^3}{\sqrt{x^2+10x+16}}=\frac{x^3}{\sqrt{(x+5)^2-9}}=\frac{(u-5)^3}{\sqrt{u^2-9}}$$

Then numerator is $$u^3-15u^2+75u-125$$. Get this as close as you can to a (scalar multiple) of $$u^2-9$$ and its derivative:

$$u^3-15u^2+75u-125=\frac12(2u)(u^2-9)-15u^2+84u-125$$

Get that quadratic leftover part as close as possible to a scalar multiple of $$u^2-9$$:

$$u^3-15u^2+75u-125=\frac12(2u)(u^2-9)-15(u^2-9)+84u-260$$

Get that linear leftover part as close as possible to a scalar multiple of $$2u$$:

$$u^3-15u^2+75u-125=\frac12(2u)(u^2-9)-15(u^2-9)+42(2u)-260$$

So you have

$$\frac12(2u)\sqrt{u^2-9}-15\sqrt{u^2-9}+42\frac{2u}{\sqrt{u^2-9}}-260\frac{1}{\sqrt{u^2-9}}$$

Each term here is relatively easy to integrate.