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
 A: 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
$$(\frac{e^x-e^{-x}}{2})^4$$
A: Well, I will give you some hints:

*

*Substitute $x=3\tan\left(\text{u}\right)$;

*For the integrand you will end up with, use the trigonometric identity:

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


*Prove the reduction formula (using integration by parts);

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


*Apply the reduction formula;

*Use the fact that:

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


*Use the fact that:

$$\sec\left(\arctan\left(x\right)\right)=\sqrt{1+x^2}\tag4$$
A: 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.
A: 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}
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
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}).
A: *

*Denominator term is to the power  -1.

*Numerator multiplied by 1.

*Use LIATE method to select first and 2nd terms for multiplication rule of Integration.

*Use multiplication rule

*Solve to get answer

