Integration over something with addition in denominator

[Xilor]

Hi,
I'm struggling to figure out how to do integration with forms such as:

∫ x/(x+1) dx
∫ x/(x+1)^2 dx
∫(b-x)^2/(b-a) dx

This last one especially is giving me a strange issue, where if I plug it into wolfram:
https://www.wolframalpha.com/input/?i=integrate+(b+-+x)^2+/(b-a)++dx

It shows up with a result of (b-x)^3/3(b-a) + C
While if we know a to be a constant 0, and is left out, and this integral is plugged in: ∫(b-x)^2/(b) dx
https://www.wolframalpha.com/input/?i=integrate+(b+-+x)^2+/(b)++dx

The answer x^3/3b +bx -x^2 +C is produced

You get a different result than the result of the former with a =0. The former one ends up with a b^2 term in the end (if you expand the (b-x)^3 and divide by b), while the latter doesn't. What is happening here and how are these types of integrals solved?

 

[jedishrfu]

Check this thread discussion

https://www.physicsforums.com/threads/how-to-integrate-x-1-x-dx.516427/

 

[Xilor]

Ah yes, that helped me do the first and second ones here, thanks! On the third I'm still completely stumped though

 

[Buffu]

∫(b-x)^2/(b-a) dx

$\displaystyle {1\over b - a}\int (b - x)^2 dx$

Now substitute $u = b -x$ or just use $(a+b)^2 = a^2 + b^2 + 2ab$.

 

[Xilor]

$\displaystyle {1\over b - a}\int (b - x)^2 dx$

Now substitute $u = b -x$ or just use $(a+b)^2 = a^2 + b^2 + 2ab$.

I assume you mean: (b−x)^2 = b^2 + x^2 - 2bx ? That's the route I tried. After integrating that part and multiplying by the 1/(b-a) again, I get:

(xb^2 + x^3/3 - bx^2)/(b-a) + C

There's no b^3 term in the numerator there, unlike the answer wolfram provided me. So it seems I'm doing something wrong (assuming wolfram isn't bugged)

 

[Mark44]

$\frac 1{b - a}\int (b - x)^2dx$

Even simpler than using a substitution, just expand the $(b - x)^2$ expression.

I assume you mean: (b−x)^2 = b^2 + x^2 - 2bx ? That's the route I tried. After integrating that part and multiplying by the 1/(b-a) again, I get:

(xb^2 + x^3/3 - bx^2)/(b-a) + C

Your answer above is fine. Possibly you entered it into wolframalpha incorrectly. Could you provide a link to the WA page with your integral?

There's no b^3 term in the numerator there, unlike the answer wolfram provided me. So it seems I'm doing something wrong (assuming wolfram isn't bugged)

 

[Xilor]

Even simpler than using a substitution, just expand the $(b - x)^2$ expression. Your answer above is fine. Possibly you entered it into wolframalpha incorrectly. Could you provide a link to the WA page with your integral?

https://www.wolframalpha.com/input/?i=integrate+(b+-+x)^2+/(b-a)++dx

My input in case that link fails:
integrate (b - x)^2 /(b-a) dx

 

[eys_physics]

You have $(xb^2+x^3/3-bx^2)/(b-a)+C=\frac{1}{3(b-a)}(3xb^2+x^3-3bx^2)+C$.
Compare this with the cubic formula $(x-b)^3=x^3+3xb^2-3bx^2-b^3$

 

[Xilor]

They are similar, but the latter has a -b^3 in there, which is not a constant. Where is this -b^3 coming from?

 

[eys_physics]

Well, $b^3$ is a constant in the sense that it is independent of $x$. You can therefore write $\frac{1}{3(b-a)}(3xb^2+x^3-3bx^2)+C=\frac{1}{3(b-a)}(x-b)^3+\frac{b^3}{3(b-a)}+C=\frac{1}{3(b-a)}(x-b)^3+D$, where $D=\frac{b^3}{3(b-a)}+C$

 

[Xilor]

I suppose so, and I guess the term would disappear if you used a definite integral. That sounds like it should be the intended reading of the result here. Thanks!

 

[Buffu]

I suppose so, and I guess the term would disappear if you used a definite integral. That sounds like it should be the intended reading of the result here. Thanks!

If you had used substitution you would have got the Wolfram answer.

$\displaystyle {1\over b - a}\int (b - x)^2 dx = {1\over a - b}\int (b-x)^2 d(b - x) = {(b-x)^3/3\over a - b}$.

 

[eys_physics]

I suppose so, and I guess the term would disappear if you used a definite integral. That sounds like it should be the intended reading of the result here. Thanks!

Yes, for an indefinite integral the constant is arbitrary. For an definite integral on the interval $[c,d]$ you would have $F(d)-F(c)$, where $F(x)$ denotes the indefinite integral. So, the constant disappears since you subtract.

 

[Buffu]

where F(x)F(x)F(x) denotes the indefinite integral

$F(x)$ denotes a antiderivative and indefinite integral is "family" of all antiderivatives. Clearly $F(x)$ does not denote a indefinite integral.

Correct me if I am wrong.