# Math Help - Integration Help

1. ## Integration Help

∫▒x^3/(1+x^2 ) dx

This is the question I have. It's quite a mind boggler (well, for me)

It has to do with further substitution with integration.

This is what I did, but I got the wrong answer in the back of the book.

Let u = 1+x^2

du/dx = 2x
du/2x = dx

∫〖(x^3/u)(du/2x)〗

∫((u-1)/2u du)

∫〖1/2-1/2u〗

=1/2 u- 1/2 ln⁡u

=1/2(1+x^2 )-1/2 ln⁡(1+x^2 )+c

I have a feeling it's something with my substitution

2. Originally Posted by macho_maggot2001
∫▒x^3/(1+x^2 ) dx

This is the question I have. It's quite a mind boggler (well, for me)

It has to do with further substitution with integration.

This is what I did, but I got the wrong answer in the back of the book.

Let u = 1+x^2

du/dx = 2x
du/2x = dx

∫〖(x^3/u)(du/2x)〗

∫((u-1)/2u du)

∫〖1/2-1/2u〗

=1/2 u- 1/2 ln⁡u

=1/2(1+x^2 )-1/2 ln⁡(1+x^2 )+c

I have a feeling it's something with my substitution
Your substitution and work is fine - when your answer is differentiated the integrand is obtained, you've included the arbitrary constant of integration, so all is well.

But remember that your answer includes an arbitrary constant (of integration) C ..... This is (probably) the clue that answers your question ......

Note that $\frac{x^3}{1 + x^2} = x - \frac{x}{1 + x^2}$. When this is integrated, you get $\frac{x^2}{2} - \frac{1}{2} \ln (1 + x^2) + K$. This answer is also correct. Is it the book's answer? How can two apparently different answers be explained?? Well ......

What happens if you replace the arbitrary constant C of your answer with the equally arbitrary $K - \frac{1}{2}$ ....?

3. Ah....

I get it now. It has to do with partial fractions. Interesting how the two are so different but have the same basic answer at the end

Thank you so much!!

4. Originally Posted by macho_maggot2001
Ah....

I get it now. It has to do with partial fractions. Interesting how the two are so different but have the same basic answer at the end

Thank you so much!!
But do you understand that your answer and the book's answer are both correct - just found using different methods. Each answer differs by a constant, so they are both correct.

5. Oh, yes I see that.

My next question has to again deal with u substitution, and I have no idea how to start it.

It is:

Integrate with respect to x:

1/((x)(x^2+16)

6. Originally Posted by macho_maggot2001
Oh, yes I see that.

My next question has to again deal with u substitution, and I have no idea how to start it.

It is:

Integrate with respect to x:

1/((x)(x^2+16)
Use partial fractions to express $\frac{1}{x(x^2 + 16)}$ in the form $\frac{A}{x} + \frac{Bx + C}{x^2 + 16}$, since $x^2 + 16$ is an irreducible quadratic. You will find that C = 0, which makes things pretty simple .....

7. Why does it have to be Bx+C

8. Originally Posted by macho_maggot2001
Why does it have to be Bx+C
Originally Posted by mr fantastic
Use partial fractions to express $\frac{1}{x(x^2 + 16)}$ in the form $\frac{A}{x} + \frac{Bx + C}{x^2 + 16}$, since $x^2 + 16$ is an irreducible quadratic. You will find that C = 0, which makes things pretty simple .....
The stuff in red is why. Are you familiar with partial fraction decomposition?

9. New questions in new topics.

Originally Posted by macho_maggot2001
Integrate with respect to x:

1/((x)(x^2+16)
Substitute $u=x^2+16,$ the rest follows easily.

10. Originally Posted by Krizalid
New questions in new topics.

Substitute $u=x^2+16,$ the rest follows easily.
The rest follows easily if the integrand is $\frac{x^2 + 16}{x}$ ...... My reading of the question is that the integrand is $\frac{1}{x(x^2 + 16)}$ .....

11. Originally Posted by mr fantastic
My reading of the question is that the integrand is $\frac{1}{x(x^2 + 16)}$ .....
That was my reading of the question too, and the rest still follows easily.

12. Originally Posted by Krizalid
That was my reading of the question too, and the rest still follows easily.
For macho_maggot2001 , I'm not sure how getting and solving $\frac{1}{2} \int \frac{1}{(u - 16) u} \, du$ is going to follow easily.

By this I mean that macho_maggot2001 might understand 'follows easily' as implying that this substitution is somehow an easier road to take than the original proposed partial fraction decomposition. Arguably, it's not.

13. We can avoid partial fraction descomposition, that's why I said "easily".

$\frac{1}
{{u(u - 16)}} = \frac{1}
{{16}} \cdot \frac{{u - (u - 16)}}
{{u(u - 16)}} = \frac{1}
{{16}}\left( {\frac{1}
{{u - 16}} - \frac{1}
{u}} \right).$

It useless to make the partial fraction descomposition since we can just tackle this with simple algebra.

14. Originally Posted by Krizalid
We can avoid partial fraction descomposition, that's why I said "easily".

$\frac{1}
{{u(u - 16)}} = \frac{1}
{{16}} \cdot \frac{{u - (u - 16)}}
{{u(u - 16)}} = \frac{1}
{{16}}\left( {\frac{1}
{{u - 16}} - \frac{1}
{u}} \right).$

It useless to make the partial fraction descomposition since we can just tackle this with simple algebra.
Don't get me wrong - I love it. And I loved your post here.

But I'm of the opinion (and only macho_maggot2001 can confirm this) that the proposed substitution and what follows is not something that would follow easily for macho_maggot2001 (even after seeing it), or something that would serve him/her well for the next similar problem .....

I just think, in context, the 'follows easily' is misleading to the member ...... My opinion, all comment welcome.

15. Oh, now I see. I believe the other post is a little easier on the eyes, but it all comes down to the same thing, right? It's all just partial fraction decomposition. I guess I didn't understand it at first because I have some inexperience with partial fractions. (I'm still in High School)

Thank you guys so much!