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

3 Answers
Dec 8, 2017

#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#

Dec 8, 2017

# 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#

Dec 8, 2017

# -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!