7
$\begingroup$

Question:
Evaluate $\displaystyle\int \frac{x^3}{\sqrt{x^2 + 4x + 6}}\ dx $.

My attempt:
$\begin{align} \int \frac{x^3}{\sqrt{x^2 + 4x + 6}}\ dx & = \int \frac{x^3}{\sqrt{(x+2)^2 + 2}}\ dx \\& \overset{(1)}= \int \frac{(\sqrt{2} \tan(t) - 2 )^3 \sqrt{2} \sec^2(t)\ }{\sqrt{2\tan^2(t) + 2}}\ dt\\& = \frac{1}{\sqrt{2}}\int\frac{(\sqrt{2} \tan(t) - 2 )^3 \sqrt{2} \sec^2(t)}{\sec(t)}\ dt\\& = \int(\sqrt{2} \tan(t) - 2 )^3 \sec(t)\ dt\\& = \int -8 \sec(t) + 2 \sqrt2 \tan^3(t) \sec(t) - 12\tan^2(t) \sec(t)\\&\qquad + 12 \sqrt2 \tan(t) \sec(t)\ dt\\\\& = -8\ln|\sec(t) + \tan(t)| + 12\sqrt{2} \sec{(t)} \\ &\qquad + 2\sqrt{2} \int \tan^3(t) \sec(t) \ dt - 12\int\tan^2(t) \sec(t) \ dt\end{align}$

Now both of these integrals can be evaluated using some sort of substitution and integration by parts rule.

This would give us, $$\boxed{-8\ln|\sec(t) + \tan(t)| + 12\sqrt{2} \sec{(t)} + 2\sqrt{2}\left[\frac{\sec^3(t)}{3} - \sec(t)\right] - 12\left[\frac{-1}{2} \ln|\sec t + \tan t| + \frac12 \sec t \tan t \right] + C}$$

$(1)$ Here I've made a substitution $t = \tan^{-1}\left(\frac{x+ 2}{\sqrt 2}\right)$.

Wolframalpha gives answer as $$\boxed{\frac{1}{3}(x^2 - 5x + 18) \sqrt{x^2 +4x + 6} - 2 \sinh^{-1}\left(\frac{x+2}{\sqrt{2}}\right) + C}$$

How would I simplify my answer equal to this? Undoing my substitution $t = \tan^{-1}\left(\frac{x+ 2}{\sqrt 2}\right)$ is also not an easy task I think.

$\endgroup$
6
  • 1
    $\begingroup$ To get the hyperbolic version, take inspiration from $\cosh^2 x = 1 + \sinh^2 x.$ $\endgroup$ Commented Jan 10, 2023 at 14:51
  • 1
    $\begingroup$ Hint: Let $x=-2+\sinh^2t$. $\endgroup$ Commented Jan 10, 2023 at 14:58
  • 2
    $\begingroup$ @owl's advice is $\sqrt{x^2+4x+6}=x+t$ so, if I've made no mistakes, $x=\frac{t^2-6}{4-2t}$ and $\int\frac{x^3dx}{\sqrt{x^2+4x+6}}=\int\frac{(t^2-6)^2dt}{8(t-2)^4}$. Then continue with this. $\endgroup$ Commented Jan 10, 2023 at 15:05
  • 6
    $\begingroup$ "Trigonometric functions don't exist" is an unhelpful hyperbolic statement. Many functions can be reduced to other functions, but there is value in using them nonetheless. $\endgroup$ Commented Jan 10, 2023 at 17:07
  • 1
    $\begingroup$ This question is yet another example (of the many that exist) where using a sinh-subs is preferable to using a tan-subs, as illustrated by Lai's solution. This dilemma, along with the one between using $x=\sec\theta$ vs $x=\cosh\theta$, is discussed here. This is one of the many weaknesses of trig/hyp substitutions that are overcome by this new technique. $\endgroup$ Commented Jun 30, 2024 at 15:08

4 Answers 4

Reset to default
5
$\begingroup$

For indefinite integral of the form $$I_n=\int \frac{x^n}{\sqrt{x^2 + bx + c}}\ dx $$ it is advised that $I_n$ be reduced first to $I_0$ before any substitution. This is achieved by the reduction formula below with $f(x)=x^2+bx+c$ $$\int \frac{f’(x)^{n}}{\sqrt{f(x)}}dx= K_n = \frac2nf’(x)^{n-1}\sqrt{f(x)}+\frac{n-1}n (b^2-4c)K_{n-2} $$ Thus, apply it to the integral to obtain \begin{align} &\int \frac{x^3}{\sqrt{x^2 + 4x + 6}}\ dx \\ =&\ \frac18\int \frac{(2x+4)^3-12(2x+4)^2+48(2x+4)-64}{\sqrt{x^2 + 4x + 6}}\ dx \\ =&\ \frac13(x^2-5x+18)\sqrt{x^2 + 4x + 6}-2\int \frac{1}{\sqrt{x^2 + 4x + 6}}\ dx \end{align}

$\endgroup$
2
$\begingroup$

Undoing the substitution is exactly how you would match the two results.

By the Pythagorean theorem, a right triangle with a reference angle $t$ such that $\tan(t)=\frac{x+2}{\sqrt2}$ has its sides occurring in a ratio of $\sqrt2$ (leg adjacent to $t$) to $x+2$ (leg opposite $t$) to $\sqrt{(x+2)^2+2}=\sqrt{x^2+4x+6}$ (hypotenuse). It follows that

$$t = \tan^{-1}\left(\frac{x+2}{\sqrt2}\right) \\ \implies \begin{cases} \tan(t) = \frac{x+2}{\sqrt2} \\ \sec(t)=\frac{\sqrt{x^2+4x+6}}{\sqrt2} \\ \sec(t)+\tan(t) = \frac{\sqrt{x^2+4x+6}+x+2}{\sqrt2} \\ \sec(t)\tan(t) = \frac{(x+2)\sqrt{x^2+4x+6}}2 \\ \ln\left|\sec(t)+\tan(t)\right| = \ln\left(\frac{\sqrt{x^2+4x+6}+x+2}{\sqrt2}\right) \end{cases}$$

After making the replacements, your antiderivative reduces to

$$-2 \ln\left(\frac{\sqrt{x^2+4x+6}+x+2}{\sqrt2}\right) + (4-3x) \sqrt{x^2+4x+6} + \frac13 (x^2+4x+6)^{3/2} + C$$

Factorize the last two terms as

$$\bigg(4-3x + \frac13 (x^2+4x+6)\bigg) \sqrt{x^2+4x+6} = -\frac13 (x^2+5x-18) \sqrt{x^2+4x+6}$$

Finally, the logarithm can be rewritten using the definition of the inverse hyp. sine:

$$\begin{align*} \ln\left(\frac{\sqrt{x^2+4x+6}+x+2}{\sqrt2}\right) &= \ln\left(\sqrt{\frac{(x+2)^2+2}2} + \frac{x+2}{\sqrt2}\right) \\ &= \ln\left(\sqrt{\left(\frac{x+2}{\sqrt2}\right)^2+1} + \frac{x+2}{\sqrt2}\right) \\[1ex] &= \sinh^{-1}\left(\frac{x+2}{\sqrt2}\right) \end{align*}$$

$\endgroup$
1
  • $\begingroup$ This is amazing! $\endgroup$ Commented Jan 11, 2023 at 17:19
2
$\begingroup$

By hyperbolic substitution

Letting $x+2=\sqrt 2 \sinh \theta$ transforms the integral into $$ \begin{aligned} I= & \int \frac{(\sqrt{2} \sinh \theta-2)^3}{\sqrt{2} \cosh \theta} \cdot \sqrt{2} \cosh \theta d \theta \\ = & \int\left(2 \sqrt{2} \sinh ^3 \theta-12 \sinh ^2 \theta+12 \sqrt{2} \sinh \theta-8\right) d \theta \\ = & 2 \sqrt{2} \int\left(\cosh ^2 \theta-1\right) d(\cosh \theta)-12 \int \frac{\cosh 2 \theta-1}{2} d \theta +12 \sqrt{2} \cosh \theta-8 \theta \\ = &\frac{2 \sqrt{2}\cosh ^3 \theta}{3}-3\sinh 2\theta +10\sqrt{2} \cosh \theta-2 \theta+C \cdots (*)\\=& \frac{2 \sqrt{2}}{3} \cosh \theta\left(15+\cosh ^2 \theta-\frac{9}{\sqrt{2}} \sinh \theta\right)-2 \theta+C \end{aligned} $$ Putting back $x+2=\sqrt 2 \sinh \theta$ yields $$ \begin{aligned} I & =\frac{2 \sqrt{2}}{3} \sqrt{1+\left(\frac{x+2}{\sqrt{2}}\right)^2}\left[15+1+\left(\frac{x+2}{\sqrt{2}}\right)^2-\frac{9}{\sqrt{2}} \cdot \frac{x+2}{\sqrt{2}}\right] +C\\ & =\frac{1}{3} \sqrt{x^2+4 x+6}\left(x^2-5 x+18\right)-2 \sinh ^{-1}\left(\frac{x+2}{\sqrt{2}}\right)+C \end{aligned} $$

By trigonometric substitution

Questioner using $x+2=\sqrt 2\tan t$ got the answer $$I=-8\ln|\sec(t) + \tan(t)| + 12\sqrt{2} \sec{(t)} + 2\sqrt{2}\left[\frac{\sec^3(t)}{3} - \sec(t)\right] - 12\left[\frac{-1}{2} \ln|\sec t + \tan t| + \frac12 \sec t \tan t \right] + C$$

Using $x+2=\sqrt2\tan t =\sqrt 2\sinh \theta \Leftrightarrow \tan t=\sinh \theta \textrm{ and } \sec t=\cosh \theta$, we get $$ \begin{aligned} I=-2 & \ln |\sec t+\tan t|+\frac{2 \sqrt{2}}{3} \sec ^3 t+10 \sqrt{2} \sec t -6 \sec t \tan t+C \\ =- & 2 \ln |\cosh \theta+\sinh \theta|+\frac{2 \sqrt{2}}{3} \cosh ^3 \theta+10\sqrt{2} \cosh \theta -6 \cosh \theta \sinh \theta+C \end{aligned} $$

Since $\ln |\cosh \theta+\sinh \theta |= \ln \left(\frac{e^\theta+e^{-\theta}}{2}+\frac{e^\theta-e^{-\theta}}{2}\right) = \theta$, therefore $$I= \frac{2 \sqrt{2}\cosh ^3 \theta}{3}-3\sinh 2\theta +10\sqrt{2} \cosh \theta-2 \theta+C \cdots (*)$$

We can now conclude that both substitutions gives the same result.

$\endgroup$
0
$\begingroup$

As you have already received an answer on how to show the $2$ results you mentioned are equivalent, I will share an alternative method to evaluate your integral.

Break the numerator as follows:

$$x^3=a(2x+4)(x^2+4x+6)+b(x^2+4x+6)+c(2x+4)+d$$ $$\implies a=\frac12,b=-6,c=5,d=4$$

So, the integral becomes

$$\int\frac{x^3}{\sqrt{x^2+4x+6}}\mathrm dx=\frac12\int \sqrt{x^2+4x+6}\mathrm d(x^2+4x+6)-6\int \sqrt{x^2+4x+6}\mathrm dx+5\int\frac{\mathrm d(x^2+4x+6)}{\sqrt{x^2+4x+6}}+4\int\frac{\mathrm dx}{\sqrt{x^2+4x+6}}$$

Evaluating all of these integrals is straightforward, albeit somewhat lengthy.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.