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

Question

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

Answer 1

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 $$(\frac{e^x-e^{-x}}{2})^4$$

Answer 2

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:

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

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

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

4. Apply the reduction formula;
5. Use the fact that:

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

6. Use the fact that:

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

Answer 3

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.

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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.

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ 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}).

Answer 5

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
5. Solve to get answer

Answer 6

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}