Anyway, the title of the webpage would be How to Solve the Integral (x+1/x+2)+3.

[mikel542]

Integration

find


(integral sign)= (x+1/x+2)+3

 

[strongmotive]

If (x+1/x+2) is one term (not sure you have written it correctly), ((x+1/x+2)^2)/2 +3x

You raise the term to the next power and divide the whole thing by that power.

 

[Dick]

If (x+1/x+2) is one term (not sure you have written it correctly), ((x+1/x+2)^2)/2 +3x

You raise the term to the next power and divide the whole thing by that power.

Dead wrong. Differentiate that and use the chain rule. It doesn't work. You have to integrate each term separately. Integrate x, 1/x, 2 and 3 and add the results.

 

[Mark44]

Integration

find


(integral sign)= (x+1/x+2)+3

What exactly is the problem? "(integral sign) = <whatever>" makes no sense to me. Is this the problem?
\int (\frac{x + 1}{x + 3} + 3)dx

Or is this it?
\int (x + 1/x + 2 + 3)dx
I suspect that this is not what you meant, although Dick interpreted what you wrote that way.

If the first integral is the one you meant, you'll need to divide (x + 1) by (x + 2), which will give you 1 + (some number)/(x + 2).

 

[icystrike]

What exactly is the problem? "(integral sign) = <whatever>" makes no sense to me. Is this the problem?
\int (\frac{x + 1}{x + 3} + 3)dx

Or is this it?
\int (x + 1/x + 2 + 3)dx
I suspect that this is not what you meant, although Dick interpreted what you wrote that way.

If the first integral is the one you meant, you'll need to divide (x + 1) by (x + 2), which will give you 1 + (some number)/(x + 2).

to thread starter :
if it is the first integral that Mark is referring to, you should most probably get 4x-ln(x+2)

 

[Dick]

to thread starter :
if it is the first integral that Mark is referring to, you should most probably get 4x-ln(x+2)

I'm not sure giving the answer is really the way we give help here, is it?