# Substitution process

1. Dec 8, 2007

### rayray19

1. The problem statement, all variables and given/known data

1) antiderivative of ((t^2)+2)/((t^3)+6t+3) dt

2) antiderivative of r(sqrt((r^2)+2))dr

2. Relevant equations

3. The attempt at a solution

#2 let u = r^2 + 2

du/dr = 2r

du = 2rdr?? i dont knoww!!

2. Dec 8, 2007

### rocomath

$$\int\frac{t^2 +2}{t^3 +6t +3}dt$$

$$\int r\sqrt{r^2+2}dr$$

correct?

3. Dec 8, 2007

### rayray19

yes, that is correct

4. Dec 8, 2007

### cristo

Staff Emeritus
For the first one, try partial fractions (probably).

For the second, your substitution seems promising. You will have the integral $$\int r u^{1/2}\frac{du}{2r}=\int \frac{1}{2}u^{1/2}du$$

Last edited: Dec 8, 2007
5. Dec 8, 2007

### rocomath

well for both, all you do is a u-substitution

so let's work 2

$$\int r\sqrt{r^2+2}dr$$

$$u=r^2 +2$$
$$du=2rdr \rightarrow \frac{1}{2}du=rdr$$

Rearranging your integral, do you notice that your derivative shows up in your original integrand? by that happening, you can take it out of your integral.

$$\int\sqrt{r^2 +1} rdr$$

$$\frac{1}{2}\int\sqrt{u}du$$

6. Dec 8, 2007

### rayray19

for my answer to #2 i got (1/3)((r^2)+2)^(3/2) + c

is that correct??

7. Dec 8, 2007

### rocomath

correct, now your first one works out the same way, all you have to do is factor our a common term from the derivative of your u-sub.

8. Dec 8, 2007

### rayray19

i factored out a 3 but now im lost at finishing it up

9. Dec 8, 2007

### rayray19

i got 3 times the antiderivative of du/u dt.. i dont think thats right though

10. Dec 8, 2007

### rocomath

$$u=t^3 +6t +3$$
$$du=3(t^2 + 1)dt \rightarrow \frac{1}{3}du=(t^2 +1)dt$$

just replace what you have with your u-sub and derivative of your u-sub.

Last edited: Dec 8, 2007
11. Dec 8, 2007

### rocomath

What you have to do is, divide by that 3 so that it becomes the constant for your substituted Integral.

Last edited: Dec 8, 2007
12. Dec 8, 2007

### rayray19

im sorry im stuck, what do i do after i find the du=3((t^2) +2)) dt

13. Dec 8, 2007

### rocomath

Last edited: Dec 8, 2007
14. Dec 8, 2007

### rayray19

so is the answer (1/3)((t^2)+2) +c ???

15. Dec 8, 2007

### rocomath

Unfortunately, no. Have you learned about the Integral of Ln? (Natural Log)

16. Dec 8, 2007

### rayray19

yea it becomes 1/x doesnt iit

17. Dec 8, 2007

### rocomath

correct, so when we complete all our substitutions, we end up with:

$$\frac{1}{3}\int\frac{1}{u}du$$

so now take the Integral of that and just resubstitute.

18. Dec 8, 2007

### rayray19

i think i have it,, (1/3)ln(t^3 + 6t + 3) ?????

19. Dec 8, 2007

### rocomath

yes, but with + C