absolutely stumped.

TimeToShine

Hi everyone

the question is the integral of root x / 1 + root x

I've let u=x^2, and plugged it back in getting 2*integral of u^2/(1+u)

When I integrate this I do u^2 seperately and 1+u seperately thus giving me u^3/3 *ln(1+u)*2

But when I plug it into wolfram it's a different answer.

Thanks

MathsManiac

TimeToShine wrote: »

Hi everyone

the question is the integral of root x / 1 + root x

I've let u=x^2, and plugged it back in getting 2*integral of u^2/(1+u)

When I integrate this I do u^2 seperately and 1+u seperately thus giving me u^3/3 *ln(1+u)*2

But when I plug it into wolfram it's a different answer.

Thanks

I think what you meant to say was that you had let u = sqrt(x), as that's what leads to your second expression.

Anyway, you can't integrate the two pieces of a product and stick the answers together like that.

Try a second substitution of letting v=1+u.

TimeToShine

alright thanks, when I do that I get 2*integral of v - 2 + 1/v

integrating that and subbing in 1+rootx for v i get

(1+rootx)^2 - 4(1+rootx) + 2ln(1+rootx) which is still different from the answer wolfram give, even after being simplified.

Michael Collins

MathsManiac wrote: »

Try a second substitution of letting v=1+u.

This should work, or else try the substitution [latex] u = 1 + \sqrt(x) [/latex] to begin with.

...Anyway, you can't integrate the two pieces of a product and stick the answers together like that.

Ah, if only eh?

TimeToShine

Thanks michael, i did that at the end but the answer's still coming out incorrectly, can't see why tbh

Michael Collins

TimeToShine wrote: »

alright thanks, when I do that I get 2*integral of v - 2 + 1/v

integrating that and subbing in 1+rootx for v i get

(1+rootx)^2 - 4(1+rootx) + 2ln(1+rootx) which is still different from the answer wolfram give, even after being simplified.

That looks right, what do you get when you simpify it? Remember that there may be an arbitrary constant in your result that Wolfram might not give you.

MathsManiac

That's correct, I believe. And wolfram is giving me: x - 2sqrt(x) + 2ln(1+sqrt(x)), which is the same thing (up to added constant).

TimeToShine

so there's supposed to be a -3 in there?

TimeToShine

well while I have you guys here might as well ask another question or two

You know when you're proving bijectivity and you get, say f(x)= a 4th or 5th degree equation. How are you supposed to express in terms of x?

for example f(x)=5x^4 + 3x^3 +2x (This is just made up)

MathsManiac

TimeToShine wrote: »

so there's supposed to be a -3 in there?

Not necessarily. When you're doing an indefinite integral, an arbitrary constant appears, because if two functions differ only by a constant, then their derivatives are the same.

The derivative of your answer and wolfram's answer are both equal to the orginal function, so they are both integrals of that function. We usually represent the arbitrary constant by including "+c" after doing an indefinite integral. Software packages usually don't do that, as they assume you understand. The only difference between your answer and wolfram's is that your "+c" and their "+c" would not be the same c.

MathsManiac

TimeToShine wrote: »

well while I have you guys here might as well ask another question or two

You know when you're proving bijectivity and you get, say f(x)= a 4th or 5th degree equation. How are you supposed to express in terms of x?

for example f(x)=5x^4 + 3x^3 +2x (This is just made up)

Not sure what you are asking. You have given f(x) in terms of x.

If it's intended to be a function whose domain and codomain are R, then it's not bijective.