Question

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

Answer 1

`int (dr)/[(r+1)*sqrt(r^2+2r)]=sec^(-1) (r+1)+C`

Explanation:

`int (dr)/[(r+1)*sqrt(r^2+2r)]`

=`int (dr)/[(r+1)*sqrt((r+1)^2-1)]`

After using `r+1=secu` and `dr=secu*tanu*du` substitution, this integral became

`int (secu*tanu*du)/[secu*sqrt((secu)^2-1)]`

=`int (tanu*du)/sqrt((tanu)^2)`

=`int (tanu*du)/(tanu`

=`int du`

=`u+c`

After using `r+1=secu` and `u=sec^(-1) (r+1)` inverse transforms, I found,

`int (dr)/[(r+1)*sqrt(r^2+2r)]=sec^(-1) (r+1)+C`

Answer 2

`int \ 1/( (r+1)sqrt(r^2+2r)) \ dr =arctan(sqrt(r^2+2r)) + C`

Explanation:

We seek:

`I = int \ 1/( (r+1)sqrt(r^2+2r)) \ dr`

If we attempt a substitution of the form:

`u = sqrt(r^2+2r) => (du)/(dr) = 1/2(r^2+2r)^(-1/2) * (2r+2)`
`:. (du)/(dr) = (r+1)/sqrt(r^2+2r) => (dr)/(du) = sqrt(r^2+2r)/(r+1)`

Substituting into the integral we get:

`I = int \ sqrt(r^2+2r)/( (r+1)(r^2+2r)) \ sqrt(r^2+2r)/(r+1) \ du`
`\ \ = int \ 1/(r+1)^2 \ du`
`\ \ = int \ 1/(r^2+2r+1) \ du`
`\ \ = int \ 1/(u^2+1) \ du`

This is now a standard integral and we have:

`I =arctan(u) + C`

And restoring the substitution we get:

`I =arctan(sqrt(r^2+2r)) + C`

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

`int \ 1/(u^2+1) \ du = arctan(u) + C`

Does require a trigonometric substitution `u=tantheta`

Answer 3

`-arc sin(1/(r+1))+C.`

Explanation:

Here is another Method to solve the Problem.

Let, `I=int1/{(r+1)sqrt(r^2+2r)}dr.`

The Substitution for this type of Integral is `(r+1)=1/x.`

`:. dr=-1/x^2dx.`

Also, `r^2+2r=(r^2+2r+1)-1=(r+1)^2-1, i.e.,`

`r^2+2r=(1/x)^2-1=(1-x^2)/x^2.`

`:. I=int1/{(1/x)sqrt((1-x^2)/x^2)}*(-1/x^2)dx,`

`=-int1/sqrt(1-x^2)dx,`

`=-arc sinx.`

Since, `(r+1)=1/x,` we have,

`I=-arc sin(1/(r+1))+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!