Evaluating $\int \frac{x^3}{\sqrt{x^2 + 4x + 6}} dx$

Question

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

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

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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.

Answer 1

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:

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

Answer 2

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}

Answer 3

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.