Integral of $~\int\frac{\sqrt{x^{2}+l^{2}}}{x}dx~$ I tried the substitution of the variable but it didn't work

Question

I need to derive the below RHS from the below LHS.

$$\int\frac{\sqrt{x^{2}+l^{2}}}{x}dx=\sqrt{x^{2}+l^{2}}-l\cdot\ln\left(\frac{l+\sqrt{x^2+l^2}}{x}\right)\tag{1}$$

$$x,l\in\mathbb R_{>0}\tag{2}$$

My tries are as below.

$$x=l\cdot\tan^{}\left(\theta_{}\right)\tag{3}$$

$$\frac{dx}{d\theta}=l\cdot\sec^{2}\left(\theta_{}\right)\tag{4}$$

$$dx=l\cdot\sec^{2}\left(\theta_{}\right)\cdot d\theta\tag{5}$$

$$\sqrt{x^{2}+l^2}=\sqrt{\left(l\cdot\tan^{}\left(\theta_{}\right)\right)^{2}+l^{2}}\tag{6}$$

$$=\sqrt{l^{2}\cdot\tan^{2}\left(\theta_{}\right)+l^{2}}\tag{7}$$

$$=\sqrt{l^{2}\left(1+\tan^{2}\left(\theta_{}\right)\right)}\tag{8}$$

$$=l\sqrt{1+\tan^{2}\left(\theta_{}\right)}\tag{9}$$

$$=l\sqrt{\sec^{2}\left(\theta_{}\right)}\tag{10}$$

$$=l\sec^{}\left(\theta_{}\right)\tag{11}$$ $$\therefore~~\int\frac{\sqrt{x^2+l^2}}{x}dx=\int\frac{l\sec^{}\left(\theta_{}\right)}{l\cdot\tan^{}\left(\theta_{}\right)}\left(l\cdot\sec^{2}\left(\theta_{}\right)\cdot d\theta\right)\tag{12}$$

$$=\int\frac{l\cdot\sec^{3}\left(\theta_{}\right)}{\tan^{}\left(\theta_{}\right)}d\theta\tag{13}$$

$$=l\int\frac{\sec^{3}\left(\theta_{}\right)}{\sin^{}\left(\theta_{}\right)\cdot\sec^{}\left(\theta_{}\right)}d\theta\tag{14}$$

$$=l\int\frac{\sec^{2}\left(\theta_{}\right)}{\sin^{}\left(\theta_{}\right)}d\theta\tag{15}$$

$$A:=\int\frac{\sec^{2}\left(\theta_{}\right)}{\sin^{}\left(\theta_{}\right)}d\theta\tag{16}$$

$$t=\sec^{}\left(\theta_{}\right)\tag{17}$$

$$\frac{dt}{d\theta}=\left(\cos^{-1}\left(\theta_{}\right)\right)'~~\leftarrow~~\text{Not a}~\arccos\left(\theta_{}\right). \tag{18}$$

$$=\left(-1\right)\left(\cos^{-2}\left(\theta_{}\right)\right)\left(-\sin^{}\left(\theta_{}\right)\right)\tag{19}$$

$$=\frac{\sin^{}\left(\theta_{}\right)}{\cos^{2}\left(\theta_{}\right)}\tag{20}$$

$$\frac{d\theta}{dt}=\frac{\cos^{2}\left(\theta_{}\right)}{\sin^{}\left(\theta_{}\right)}\tag{21}$$

$$d\theta=\left(\frac{\cos^{2}\left(\theta_{}\right)}{\sin^{}\left(\theta_{}\right)}\right)dt\tag{22}$$

$$\cos^{}\left(\theta_{}\right)=\sec^{-1}\left(\theta_{}\right)\tag{23}$$

$$\therefore~~\cos^{2}\left(\theta_{}\right)=\sec^{-2}\left(\theta_{}\right)\tag{24}$$

$$d\theta=\left(\frac{\sec^{-2}\left(\theta_{}\right)}{\sin^{}\left(\theta_{}\right)}\right)dt\tag{25}$$

$$A=\int\frac{\sec^{2}\left(\theta_{}\right)}{\sin^{}\left(\theta_{}\right)}d\theta=\int\frac{\sec^{2}\left(\theta_{}\right)}{\sin^{}\left(\theta_{}\right)}\left(\left(\frac{\sec^{-2}\left(\theta_{}\right)}{\sin^{}\left(\theta_{}\right)}\right)dt\right)\tag{26}$$

$$=\int\frac{1}{\sin^{2}\left(\theta_{}\right)}dt~~\leftarrow~~\text{Not good}\tag{27}$$

How should I've done of the integration by substitution?

Answer 1

$ \newcommand{\t}{\theta} \newcommand{\d}{\mathrm{d}} \newcommand{\ds}{\displaystyle} \newcommand{\I}{\mathcal{I}} $ Begin with

$$A = \int \frac{\sec^2 \t}{\sin \t} \, \d \t$$

Use the Pythagorean identity $\sec^2 \t = 1 + \tan^2 \t$. Then with some algebra,

$$A = \underbrace{\int \csc \t \, \d \t}_{\ds =: \I_1} + \underbrace{\int \frac{\sin \t}{\cos^2 \t} \, \d \t}_{\ds =: \I_2}$$

$\I_2$ is trivial enough with the $u$-substitution $u := \cos \t$, which would give $\d u = - \sin \t \, \d \t$, so

$$\I_2 = - \int u^{-2} \, \d u = \frac{1}{u} + C = \frac{1}{\cos \t} +C = \sec \t + C$$

The first is a bit less trivial, but if you're familiar with the technique for solving $\int \sec(x) \, \d x$, it's quite similar. We have

$$\I_1 = \int \csc \t \frac{\csc \t + \cot \t}{\csc \t + \cot \t} \, \d \t$$

Use $u := \csc \t + \cot \t$; then $\d u = (- \csc^2 \t - \csc\t \cot \t) \, \d \t$, i.e. $-1$ times the numerator:

$$\I_1 = - \int \frac{1}{u} \, \d u = - \ln |u| + C = - \ln \big| \csc \t + \cot \t \big| + C$$

Therefore,

$$A = \sec \t - \ln \big| \csc \t + \cot \t \big| + C$$

Note that you had $\t = \arctan(x/\ell)$ and you can finish from here; what remains is basic trig and algebra.

Answer 2

There's also a very much simpler way .

In $I = \int \frac{ \sqrt{ x ^{2} + l ^{2} } }{ x } dx $.

Multiply and divide by $\sqrt{x^2 + l^2} $,we get ,

$$I = \int {\frac{x}{\sqrt{{x^2} + {l^2}}} }dx + {l^2} \int \frac {dx}{x \sqrt {x^2 + l^2}} $$

Put $x^2 + l^2 = t^2$ into first term and $x=l \tan{\theta}$ in the second term ,we get

$$I = t + \ln { |\tan { \theta \over 2} | } + C $$ .

Now put the values and get the answer .

Answer 3

Let $x=\ell\tan\theta$ results in $$ \int\frac{\sqrt{x^2+\ell^2}}x\,\mathrm{d}x= \ell\int\frac{\sec^3\theta}{\tan\theta}\,\mathrm{d}\theta=\ell\int\frac{\sin\theta\,\mathrm{d}\theta}{\cos^2\theta(1-\cos^2\theta)} $$ So letting $c=\cos\theta$ gives $$ \dots=-\ell\int\frac{\mathrm{d}c}{c^2(1-c^2)}=-\ell\int\left(\frac1{c^2}+\frac1{1-c^2}\right)\,\mathrm{d}c $$ which you should know how to integrate.

Alternatively, let $x=\ell\sinh t$ gives $$ \int\frac{\sqrt{x^2+\ell^2}}x\,\mathrm{d}x= \ell\int\frac{\cosh^2 t}{\sinh t}\,\mathrm{d}t= \ell\int\left(\sinh t+\frac1{\sinh t}\right)\,\mathrm{d}t $$ etc.

Or you can let $x=\ell\cdot\frac{2t}{1-t^2}$, etc.

Answer 4

In this answer, I wanted to show that it is possible to use an algebraic substitution.

Take $$u=x+\sqrt{x^2+l^2}$$ then we have,

$$\begin{align}&u-x=\sqrt{x^2+l^2} \\ \implies &x^2+l^2=(u-x)^2\\ \implies &l^2=u^2-2ux \\ \implies &x=\frac{u^2-l^2}{2u}\end{align}$$

and

$$\mathrm{d}x=\frac{l^2+u^2}{2u^2}$$

Finally,

$$\begin{align}\int\frac{\sqrt{x^{2}+l^{2}}}{x}\mathrm {d}x=\frac 12 \int \frac{(u^2+l^2)^2}{u^2(u^2-l^2)}\mathrm{d}u=\frac 12 \int \left(-\frac{l^2}{u^2} + \frac{2 l}{u-l} - \frac{2 l}{u+l} + 1\right) \mathrm{d}u\end{align}$$

Answer 5

*Another way to do it.$

Let $$u=\sqrt{x^2+l^2}\implies du=\frac{x}{\sqrt{x^2+l^2}}dx$$ $$\int\frac{\sqrt{x^{2}+l^{2}}}{x}dx=\int\frac{u^2}{u^2-l^2}\,du=\int\frac{u^2-l^2+l^2}{u^2-l^2}\,du=\int du+l ^2 \int\frac{du}{(u+l)(u-l)}$$ Using partial fraction decomposition $$\frac{1}{(u+l)(u-l)}=\frac 1{2l} \left(\frac 1{u-l}-\frac 1{u+l} \right)$$ which becomes to be quite simple.

When back to $x$, simplify the logarithms.