# Homework Help: Integration exam question

1. Aug 10, 2011

### Shannabel

1. The problem statement, all variables and given/known data
find the integral from 0 to 3 of (5x^2+30)/(x^4+13x^2+36)

2. Relevant equations

3. The attempt at a solution
I thought it may be an integration by parts question, so I tried that first, but the book says that I should be using substitution.
So I tried using u=x^2:
du=2xdx
x=u^(1/2)

(5u+30)du/(u^2+13u+36)(2u^(1/2))
so now I have
(5u+30)du/(2u^(5/2)+26u^(3/2)+72u^(1/2))

but i have no better idea how to integrate that than i did the original equation. any ideas?

2. Aug 10, 2011

### Bohrok

This problem would probably be best done by partial fraction decomposition first, then substitutions.

3. Aug 10, 2011

### Staff: Mentor

I agree with Bohrok.

4. Aug 10, 2011

### Shannabel

do you mean using something like
(Ax+B)/(x^2+4)+(Cx+D)/(x^2+9)=5x^2+30?

5. Aug 10, 2011

### Staff: Mentor

Not quite.
It would be

(Ax+B)/(x^2+4) + (Cx+D)/(x^2+9) = (5x^2+30)/[(x^2+4)(x^2+9)]

The goal is to solve for A, B, C, and D so that this equation is identically true.

6. Aug 10, 2011

### Shannabel

okay, so i solved and got
A=0
B=3
C=0
D=2

3/(x^2+4)+2/(x^2+9)

which i could integrate to get
3ln(x^2+4)+2ln(x^2+9)

and i could continue, but my possible answers all involve arctan so i think i'm on the wrong track here...?

7. Aug 10, 2011

### Staff: Mentor

I didn't check your values for A, B, etc.

Your integrals are incorrect. You should not be getting log expressions, but you should be getting arctan expressions.

8. Aug 10, 2011

### Staff: Mentor

Also, check your work for the constants.
3/(x^2 + 4) + 2/(x^2 + 9) = (5x^2 + 35)/[(x^2 + 4)(x^2 + 9)] $\neq$ (5x^2 + 30)/[(x^2 + 4)(x^2 + 9)]

9. Aug 10, 2011

### Shannabel

how do i get arctan expressions? because the derivative of tan is 1/(a^2+x^2), and i have 2 and 3 rather than 1 as my numerators?

10. Aug 10, 2011

### Shannabel

sorry..
3/(x^2+9)+2/(x^2+4) ,right?

11. Aug 10, 2011

### Staff: Mentor

No, d/dx(tan(x)) = sec2(x).

d/dx(tan-1(x)) = 1/(x2 + 1) - that must be what you're thinking about.

Also, d/dx(tan-1(x/a)) = (1/a)/((x/a)2 + 1)
This implies that ∫(1/a)/((x/a)2 + 1) dx = tan-1(x/a) + C

12. Aug 10, 2011

### Shannabel

i'm confused....

13. Aug 10, 2011

### Bohrok

If d/dx (x2) = 2x, then ∫2x dx = x2 + C by integrating both sides. Mark just did the same thing with tan-1(x/a) where a is a constant.

14. Aug 10, 2011

### Shannabel

okay i think that makes sense... but how can i use that to solve the problem?

15. Aug 10, 2011

### Bohrok

That'll come in later; first check over your work for the constants in the numerators of the partial fractions. The numbers are almost correct.

16. Aug 10, 2011

### Shannabel

(ax+b)(x^2+4)+(cx+d)(x^2+9)=(5x^2+30)
ax^3+bx^2+4Ax+4B+Cx^3+9Cx+Dx^2+9D=5X62+30

A+C=0 and 4A+9C=0 therefore A and C both equal 0
B+D=5 and 4B+9D=30, so 5D=10 and D=2 so B=3
and i have
3/(x^2+9)+2(x^2+4)

:( can't figure out what's wrong!

17. Aug 10, 2011

### vela

Staff Emeritus
Earlier, you wrote
which wasn't correct. Now, you have
which is correct. You swapped the coefficients earlier.

18. Aug 10, 2011

### Staff: Mentor

This is correct. You can easily check by verifying that 3/(x^2+9)+2/(x^2+4) = (5x^2 + 30)/[(x^2+9)(x^2+4)]

Last edited: Aug 10, 2011
19. Aug 10, 2011

### Shannabel

okay, so from here i should integrate and end up with arctan functions... how do i do this?

20. Aug 10, 2011

### Staff: Mentor

$$\int \frac{a~dx}{x^2 + a^2} = tan^{-1}(x/a) + C$$

This formula is essentially the same as the one I gave in an earlier post.

A mistake you made earlier was thinking that the following was true:
$$\int \frac{dx}{x^2 + a^2} = ln(x^2 + a^2) + C$$

More generally, this also is not true:
$$\int \frac{dx}{f(x)} = ln(f(x)) + C$$

21. Aug 10, 2011

### SammyS

Staff Emeritus
This should be: 3/(x^2+9)+2/(x^2+4).

To integrate the first term:

$$\int\frac{3\,dx}{x^2+9}=\int\frac{3\,dx}{9((x/3)^2+1)}$$
$\displaystyle =\int\frac{dx}{3((x/3)^2+1)}$​
Let u=x/3 → du=(1/3)dx

The integral becomes: $\displaystyle \int\frac{du}{u^2+1}=\tan^{-1}(u) + C$

22. Aug 10, 2011

### Shannabel

Got it! Thanks :)

23. Aug 10, 2011

### Staff: Mentor

Good eye - I missed that.