# I Integration over something with addition in denominator

1. Apr 6, 2017

### 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?

2. Apr 6, 2017

3. Apr 6, 2017

### Xilor

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

4. Apr 6, 2017

### Buffu

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

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

5. Apr 6, 2017

### Xilor

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)

6. Apr 6, 2017

### Staff: Mentor

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

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?

7. Apr 6, 2017

### Xilor

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

8. Apr 6, 2017

### 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 $(x-b)^3=x^3+3xb^2-3bx^2-b^3$

9. Apr 6, 2017

### 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?

10. Apr 6, 2017

### 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 $D=\frac{b^3}{3(b-a)}+C$

11. Apr 6, 2017

### 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!

12. Apr 6, 2017

### Buffu

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

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

13. Apr 6, 2017

### eys_physics

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.

14. Apr 7, 2017

### Buffu

$F(x)$ 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.