# Method to solve this indefinite integral $\int \frac{x^{4}}{\sqrt{x^{2}+9}}dx$

I'm not able to solve this integral. Can anybody help me? I tried doing it by parts noticing that $$\int \frac{x^{4}}{\sqrt{x^{2}+9}} \, dx \, = \int x^{3} \cdot \frac{x}{\sqrt{x^{2}+9}} .$$ where we know that $$D[\frac{x}{\sqrt{x^{2}+9}}] = \sqrt{x^{2}+9}$$

so that we can view the integral as

$$\int \frac{x^{4}}{\sqrt{x^{2}+9}} \, dx \, = x^{3} \cdot \sqrt{x^{2}+9} - \int 3 \cdot x^{2} \cdot \sqrt{x^{2}+9} \, dx .$$ which led me in another integration by parts and in another one again, so i realized that maybe this wasn't the right way to do it

• OK, what have you tried? Dec 28, 2020 at 20:21
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– Joe
Dec 28, 2020 at 20:23
• oh, i'm sorry. I'm new to the forum. I should've add the methods i used and my difficulties related to this problem.
– Raff
Dec 28, 2020 at 20:43
• @Raff You can still do that by editing your question. See my previous comment.
– Joe
Dec 28, 2020 at 20:48
• I guess that is a bit more settled as question, though i have another question. Was my method completely hopeless?
– Raff
Dec 28, 2020 at 21:10

hint

We know that $$(\forall t\in \Bbb R)\;\; \sqrt{1+\sinh^2(t)}=\cosh(t)$$

and

$$\sqrt{9+(3\sinh(t))^2}=3\cosh(t)$$ So, with $$t=3\sinh(t)$$,

the integral becomes

$$81\int \sinh^4(t)dt$$

now, use linearization by expanding $$(\frac{e^x-e^{-x}}{2})^4$$

• Thanks @hamam_Abdallah
– Raff
Dec 28, 2020 at 20:55
• @Raff You are Welcome and happy new year without covid. Dec 28, 2020 at 20:57

Well, I will give you some hints:

1. Substitute $$x=3\tan\left(\text{u}\right)$$;
2. For the integrand you will end up with, use the trigonometric identity:

$$\tan^2\left(x\right)=\sec^2\left(x\right)-1\tag1$$

1. Prove the reduction formula (using integration by parts);

$$\int\sec^\text{n}\left(x\right)\space\text{d}x=\frac{\sin\left(x\right)\sec^{\text{n}-1}\left(x\right)}{\text{n}-1}+\frac{\text{n}-2}{\text{n}-1}\int\sec^{\text{n}-2}\left(x\right)\space\text{d}x\tag2$$

1. Apply the reduction formula;
2. Use the fact that:

$$\int\sec\left(x\right)\space\text{d}x=\ln\left|\tan\left(x\right)+\sec\left(x\right)\right|+\text{C}\tag3$$

1. Use the fact that:

$$\sec\left(\arctan\left(x\right)\right)=\sqrt{1+x^2}\tag4$$

• So the methods to solve this integral are using an recursive formula for an integral or the hyperbloic substitution. The problem is that my course doesn't study deep this method so are a bit obscure to me. the only thing that i've tried was doing it by parts. But if i want be able to solve any integral i'll have to go beyond my course. Thanks @Jan Eerland
– Raff
Dec 28, 2020 at 20:55

Since other answerers have already explained how to solve the question at hand, I'll focus on a more general problem solving strategy. The integral in question, $$\int \frac{x^4}{\sqrt{x^2+9}} \, dx \, ,$$ has an expression involving $$\sqrt{x^2+9}$$. This should make it obvious that a trigonometric substitution is the way to go, and I'll explain why. The identities \begin{align} \sin(\theta)^2+\cos(\theta)^2&=1 \tag{1}\label{1}\\ \tan(\theta)^2+1&=\sec(\theta)^2 \tag{2}\label{2} \end{align} come in handy here. $$\eqref{1}$$ can be rearranged to $$\cos(\theta)=\sqrt{1-\sin(\theta)^2} \, .$$ (The reason that we don't bother writing $$\cos(\theta)=\pm\sqrt{1-\sin(\theta)^2}$$ is quite subtle; I'll leave it as a footnote if you're interested.)

These identities imply that any integral of the form $$\int f\left(\sqrt{1-x^2}\right) \, dx$$ can be simplified if we let $$x=\sin(\theta)$$. Since $$dx=\cos(\theta)d\theta$$, we obtain $$\int f(\cos(\theta))\cos(\theta) \, d\theta \, ,$$ meaning we can avoid the pesky square root sign. For much the same reason, $$\int f(x^2+1) \, dx$$ can often be simplified if we write $$x=\tan(\theta)$$. Your problem was almost the same—it is just that we had to substitute $$x=3\tan(\theta)$$ because we were dealing with $$\sqrt{x^2+9}$$ rather than $$\sqrt{x^2+1}$$. Integration by parts doesn't offer an obvious way of simplifying the integral, unlike this method.

To fully appreciate why we don't write $$\cos(\theta)=\sqrt{1-\sin(\theta)^2}$$ one must be familiar with why integration by substitution works in the first place:

Integration by substitution

Integration by substitution comes from reversing the chain rule. Recall that if $$y=f(g(x))$$, then $$\frac{dy}{dx}=f'(g(x))g'(x) \, .$$ Thus, $$\int f'(g(x))g'(x) \, dx = f(g(x))+C \, .$$ On the other hand, if we make the substitutions $$u=g(x)$$ and $$du = g'(x) \, dx$$, then the integral becomes $$\int f'(u) \, du=f(u)+C=f(g(x))+C \, .$$ It just so happens that working out $$du/dx$$ and then 'multiplying by $$du$$' means that you will make the correct substitutions. The formula can be summarised as $$\int f(g(x))g'(x) \, dx = \int f'(u) \, du \quad \text{where u=g(x)} \, .$$ In Leibnizian notation, it reads particularly well $$\int \frac{dy}{du} \cdot \frac{du}{dx} \, dx = \int \frac{dy}{du} \, du \, .$$ Again, the $$dx$$'s 'cancel' in a similar fashion to the chain rule.

More Sophisticated Substitutions

A more sophisticated way of making substitutions is given in Michael Spivak's Calculus. Integrals of the form $$\int f'(g(x))g'(x) \, dx$$ can be easily solved by substituting $$u=g(x)$$. In fact, many of these integrals are so simple that you can do them in your head. However, the substitution $$u=g(x)$$ can still be made even if the factor $$g'(x)$$ does not appear. In general, $$\int f(g(x)) \, dx \tag{*}\label{*}$$ can be solved in the following way, provided that $$g$$ is one-to-one: \begin{align} u &= g(x) \\ x &= g^{-1}(u) \\ \frac{dx}{du} &= (g^{-1})'(u) \\ dx &= (g^{-1})'(u)du \end{align} This means that \eqref{*} can be transformed to $$\int f(u)(g^{-1})'(u) \, du \, ,$$ which in practice often makes the integral simpler to solve than before. Hence, $$\int f(g(x)) \, dx = \int f(u)(g^{-1})'(u) \, du \, .$$ The validity of this approach can be demonstrated by differentiating $$\int f(u)(g^{-1})'(u)$$ with respect to $$x$$: \begin{align} \frac{d}{dx}\int f(u)(g^{-1})'(u) \, du &= f(u)(g^{-1})'(u)\frac{du}{dx} \\ &= f(u) \frac{dx}{du} \frac{du}{dx} \\ &= f(g(x)) \, . \end{align}

Why we ignore the $$\pm$$ signs

This means that we can make a substitution of the form $$x=\ldots$$, what we are really doing is making a substitution of the form $$u=\ldots$$ and then taking the inverse. Say we are solving $$\int \sqrt{1-x^2} \, dx \, .$$ When we write $$\text{let x=\sin(\theta)} \, ,$$ what we really mean is let $$\theta=\arcsin(x)$$, meaning that $$x=\sin(\theta)$$ with $$-\pi/2 \leq \theta \leq \pi/2$$. Notice that $$\arcsin$$ is a one-to-one function, meaning that the substitution is valid. Then, since $$-\pi/2 < \theta < \pi/2$$, we know that $$\cos(\theta)$$ is positive, so we can drop the $$\pm$$ sign.

• (The reason that we don't bother writing cos(θ)=±sqrt(1−sin(θ)^2) is quite subtle; I'll leave it as a footnote if you're interested.). I'd love knowing this information
– Raff
Dec 29, 2020 at 14:38
• @Raff OK, I will update my answer. It will take some time for me to get to the question at hand, but I'll get there in the end. If you have any questions, please ask.
– Joe
Dec 29, 2020 at 16:02
• thank you very much, I really appreciate it
– Raff
Dec 30, 2020 at 17:33

Note that $$\int \frac{x^{4}}{\sqrt{x^{2}+9}} \, \overset{x=3t}{dx} =81 \int \frac{t^{4}}{\sqrt{t^{2}+1}} dt$$ and apply the following reduction formula twice $$\int\frac{t^n}{\sqrt{t^2+1}}dt=I_n=\frac1n t^{n-1}\sqrt{t^2+1}-\frac{n-1}nI_{n-2}$$ to obtain \begin{align} \int \frac{t^{4}}{\sqrt{t^{2}+1}} dt =\frac14t^3\sqrt{t^2+1}-\frac38t \sqrt{t^2+1}+\frac38\sinh^{-1}t \end{align}

$$\newcommand{\bbx}{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}{\mathcal{#1}} \newcommand{\mrm}{\mathrm{#1}} \newcommand{\on}{\operatorname{#1}} \newcommand{\pars}{\left(\,{#1}\,\right)} \newcommand{\partiald}[]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}{\left\vert\,{#1}\,\right\vert}$$ With $$\ds{\bbox[5px,#ffd]{x \equiv {9 - t^{2} \over 2t} \implies t = \root{x^{2} + 9} - x}}$$: \begin{align} &\bbox[5px,#ffd]{\int{x^{4} \over \root{x^{2} + 9}} \,\dd x} \\[5mm] = &\ \int\pars{-\,{t^{3} \over 16} + {9t \over 4} - {243 \over 8t} + {729 \over 4t^{3}} - {6561 \over 16t^{5}}}\dd t \\[5mm] = &\ -\,{t^{4} \over 64} + {9t^{2} \over 8} - {243\ln\pars{t} \over 8} - {729 \over 8t^{2}} + {6561 \over 64t^{4}} + \mbox{a constant}\label{1}\tag{1} \end{align} Replace $$\ds{\quad t = \root{x^{2} + 9} - x\quad}$$ in (\ref{1}).

1. Denominator term is to the power -1.
2. Numerator multiplied by 1.
3. Use LIATE method to select first and 2nd terms for multiplication rule of Integration.
4. Use multiplication rule