3
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ No. How did you eliminate the square root in the first step? By the way, you should get: $$\frac{1}{6} \sqrt{x^2+10 x+16} \left(2 x^2-25 x+311\right)-\frac{385}{2} \log \left(\sqrt{x^2+10 x+16}+x+5\right))$$ $\endgroup$ Commented Dec 24, 2019 at 1:59
  • $\begingroup$ It's a typo. I edited it. $\endgroup$
    – romeoPH
    Commented Dec 24, 2019 at 2:20
  • $\begingroup$ @DavidG.Stork, he is getting this only, if you combine first two terms. $\endgroup$
    – Martund
    Commented Dec 24, 2019 at 2:38

3 Answers 3

9
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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}} $

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .