Solving Substitution Process Homework Questions

[rayray19]

Homework Statement

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



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



help please with these

Homework Equations

The Attempt at a Solution

#2 let u = r^2 + 2

du/dr = 2r

du = 2rdr?? i don't knoww!

 

[rocomath]

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

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

correct?

 

[rayray19]

yes, that is correct

 

[cristo]

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

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

 

[rocomath]

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

1) u-sub of your denominator

2) u-sub of your radican, which you already did

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.

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

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

 

[rayray19]

for my answer to #2 i got (1/3)((r^2)+2)^(3/2) + cis that correct??

 

[rocomath]

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


is that correct??

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.

 

[rayray19]

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

 

[rayray19]

i got 3 times the antiderivative of du/u dt.. i don't think that's right though

 

[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.

 

[rocomath]

i got 3 times the antiderivative of du/u dt.. i don't think that's right though

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

 

[rayray19]

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

 

[rocomath]

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

scroll up to post #10 and then look at the last part of post #5 to see how i arranged my substituted Integral.

https://www.physicsforums.com/showpost.php?p=1534291&postcount=10

https://www.physicsforums.com/showpost.php?p=1534278&postcount=5

 

[rayray19]

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

 

[rocomath]

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

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

 

[rayray19]

yea it becomes 1/x doesn't iit

 

[rocomath]

yea it becomes 1/x doesn't iit

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.

 

[rayray19]

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

 

[rocomath]

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

yes, but with + C

you can always confirm your answer by taking the derivative, gl!

 

[cristo]

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

It helps if you post your working at each stage, instead of simply posting what you get as the answer. This not only helps the person checking your work, but also helps you in that you organise your thoughts into a logical progression through the problem.

 

[rayray19]

thank yo u so muchh, I am working on a problem now

antiderivative of 4rsqrt(8-r)dr

so the u is 8-r
and the du is -1dr right

 

[rocomath]

yes, that's correct

 

[rayray19]

now i got up to 4 times antiderivative of (8-u)sqrt(u)) -du

right?

 

[rayray19]

antiderivative of 4rsqrt(8-r)dr

my answer is...

(8/5)(8-r)^(5/2) - (64/3)(8-r)^(3/2) + ccan any1 tell me if this is correct

 

[coomast]

antiderivative of 4rsqrt(8-r)dr

my answer is...

(8/5)(8-r)^(5/2) - (64/3)(8-r)^(3/2) + c


can any1 tell me if this is correct

It is correct.