Integration via substitution problem.

In summary, the conversation is about a question on integrating using substitution, specifically the problem of x/(x+1)^0.5 dx. The person asking the question has followed the substitution method and arrived at their own answer, but it differs from the answer given in the book. However, upon further examination, it is revealed that the two answers are actually equivalent, with the book's version being a simplified version of the person's answer. The conversation ends with the person thanking for spotting their mistake and apologizing for the confusion.
  • #1
Gaz031
51
0
Hi, I'm currently stuck on an integration via substitution problem. I have an answer but the one given in the book of the book is different to mine. I'm wondering where exactly I've gone wrong, if i have:

Q10: Integrate:

x/ (x+1)^0.5 dx. Use the substitution, u^2 = x + 1.

Heres my working:
u^2 = x + 1.
u = (x+1)^0.5
2u(du/dx) = 1
x = u^2 - 1

So, using some substitution:

(u^2 - 1)/u 1dx
(u^2 - 1)/u 2u(du/dx)dx
(u^2 - 1)2 du
(2u^2 - 2) du

Now integrating with respect to u:

(2/3)u^3 - 2u

Substituting u = (x+1)^0.5
(2/3).(x+1)^1.5 - 2.(x+1)^0.5

However, the actual answer given in the back of the book is:

(2/3)(x-2).(x+1)^0.5

Could anyone spot my mistake for me? Thanks a lot.
 
Physics news on Phys.org
  • #2
You haven't made a mistake. (2/3)(x + 1)^1.5 - 2(x + 1)^0.5 = (x + 1)^0.5( (2/3)(x + 1)^1 - 2) = (2/3 * (x - 1)) * (x + 1)^0.5, i.e what the book wrote. Also, don't forget about the constant of integration.
 
  • #3
Ooops. Their version is just simplified. Trust me to get the part that was new to me right then forget to simplify with basic algebra >_<.
Thanks, sorry for the stupid topic.
 

FAQ: Integration via substitution problem.

What is integration via substitution?

Integration via substitution is a technique used in calculus to solve integrals by substituting a variable with a different expression. This allows for a simpler integral to be solved.

How do I know when to use integration via substitution?

You can use integration via substitution for integrals that involve a product of functions, a composite function, or a function raised to a power. Look for expressions that can be substituted with a single variable.

What is the general process for solving an integral using substitution?

The general process for integration via substitution involves identifying a substitution that will simplify the integral, substituting the expression and its derivative into the integral, and then solving the resulting integral using basic integration techniques.

Can integration via substitution be used for all types of integrals?

No, integration via substitution is not always the best method for solving integrals. It is most useful for integrals that involve a product of functions or a composite function. Other methods, such as integration by parts, may be more effective for other types of integrals.

Are there any tips for choosing the best substitution for an integral?

Some tips for choosing the best substitution for an integral include looking for expressions that can be simplified or combined, and choosing a substitution that will result in a simpler integral. It may also be helpful to try multiple substitutions to see which one works best.

Back
Top