# Evaluate the integral? : int 1/( (r+1)sqrt(r^2+2r)) dr

Dec 8, 2017

$\int \frac{\mathrm{dr}}{\left(r + 1\right) \cdot \sqrt{{r}^{2} + 2 r}} = {\sec}^{- 1} \left(r + 1\right) + C$

#### Explanation:

$\int \frac{\mathrm{dr}}{\left(r + 1\right) \cdot \sqrt{{r}^{2} + 2 r}}$

=$\int \frac{\mathrm{dr}}{\left(r + 1\right) \cdot \sqrt{{\left(r + 1\right)}^{2} - 1}}$

After using $r + 1 = \sec u$ and $\mathrm{dr} = \sec u \cdot \tan u \cdot \mathrm{du}$ substitution, this integral became

$\int \frac{\sec u \cdot \tan u \cdot \mathrm{du}}{\sec u \cdot \sqrt{{\left(\sec u\right)}^{2} - 1}}$

=$\int \frac{\tan u \cdot \mathrm{du}}{\sqrt{{\left(\tan u\right)}^{2}}}$

=int (tanu*du)/(tanu

=$\int \mathrm{du}$

=$u + c$

After using $r + 1 = \sec u$ and $u = {\sec}^{- 1} \left(r + 1\right)$ inverse transforms, I found,

$\int \frac{\mathrm{dr}}{\left(r + 1\right) \cdot \sqrt{{r}^{2} + 2 r}} = {\sec}^{- 1} \left(r + 1\right) + C$

Dec 8, 2017

$\int \setminus \frac{1}{\left(r + 1\right) \sqrt{{r}^{2} + 2 r}} \setminus \mathrm{dr} = \arctan \left(\sqrt{{r}^{2} + 2 r}\right) + C$

#### Explanation:

We seek:

$I = \int \setminus \frac{1}{\left(r + 1\right) \sqrt{{r}^{2} + 2 r}} \setminus \mathrm{dr}$

If we attempt a substitution of the form:

$u = \sqrt{{r}^{2} + 2 r} \implies \frac{\mathrm{du}}{\mathrm{dr}} = \frac{1}{2} {\left({r}^{2} + 2 r\right)}^{- \frac{1}{2}} \cdot \left(2 r + 2\right)$
$\therefore \frac{\mathrm{du}}{\mathrm{dr}} = \frac{r + 1}{\sqrt{{r}^{2} + 2 r}} \implies \frac{\mathrm{dr}}{\mathrm{du}} = \frac{\sqrt{{r}^{2} + 2 r}}{r + 1}$

Substituting into the integral we get:

$I = \int \setminus \frac{\sqrt{{r}^{2} + 2 r}}{\left(r + 1\right) \left({r}^{2} + 2 r\right)} \setminus \frac{\sqrt{{r}^{2} + 2 r}}{r + 1} \setminus \mathrm{du}$
$\setminus \setminus = \int \setminus \frac{1}{r + 1} ^ 2 \setminus \mathrm{du}$
$\setminus \setminus = \int \setminus \frac{1}{{r}^{2} + 2 r + 1} \setminus \mathrm{du}$
$\setminus \setminus = \int \setminus \frac{1}{{u}^{2} + 1} \setminus \mathrm{du}$

This is now a standard integral and we have:

$I = \arctan \left(u\right) + C$

And restoring the substitution we get:

$I = \arctan \left(\sqrt{{r}^{2} + 2 r}\right) + C$

Note that although this does not explicitly use a trigonometric substitution that the derivation of the standard result

$\int \setminus \frac{1}{{u}^{2} + 1} \setminus \mathrm{du} = \arctan \left(u\right) + C$

Does require a trigonometric substitution $u = \tan \theta$

Dec 8, 2017

$- a r c \sin \left(\frac{1}{r + 1}\right) + C .$

#### Explanation:

Here is another Method to solve the Problem.

Let, $I = \int \frac{1}{\left(r + 1\right) \sqrt{{r}^{2} + 2 r}} \mathrm{dr} .$

The Substitution for this type of Integral is $\left(r + 1\right) = \frac{1}{x} .$

$\therefore \mathrm{dr} = - \frac{1}{x} ^ 2 \mathrm{dx} .$

Also, ${r}^{2} + 2 r = \left({r}^{2} + 2 r + 1\right) - 1 = {\left(r + 1\right)}^{2} - 1 , i . e . ,$

${r}^{2} + 2 r = {\left(\frac{1}{x}\right)}^{2} - 1 = \frac{1 - {x}^{2}}{x} ^ 2.$

$\therefore I = \int \frac{1}{\left(\frac{1}{x}\right) \sqrt{\frac{1 - {x}^{2}}{x} ^ 2}} \cdot \left(- \frac{1}{x} ^ 2\right) \mathrm{dx} ,$

$= - \int \frac{1}{\sqrt{1 - {x}^{2}}} \mathrm{dx} ,$

$= - a r c \sin x .$

Since, $\left(r + 1\right) = \frac{1}{x} ,$ we have,

$I = - a r c \sin \left(\frac{1}{r + 1}\right) + C .$

Enjoy Maths.!

N.B. : I leave it as an Exercise to the Questioner to

show that the other 2 Answers obtained by Respected

Cem Sentin and Steve M are the same!