Indefinite integrals

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January 9th 2010, 06:38 AM
mj.alawami

Indefinite integrals

Question:
$\int \sqrt[3] {3x+5}.dx$

So the equation will be $\int (3x+5)^\frac{1}{3}.dx$

$u=3x+5 \rightarrow \int u^\frac{1}{3}$
$du=3$
After this step i dont know what to do , do i divide by 3 to produce 1?
Please answer in step .Thank you
January 9th 2010, 06:42 AM
HallsofIvy

Quote:

Originally Posted by mj.alawami

Question:
$\int \sqrt[3] {3x+5}.dx$

So the equation will be $\int (3x+5)^\frac{1}{3}.dx$

$u=3x+5 \rightarrow \int u^\frac{1}{3}$
$du=3$

Here's your problem! du is NOT equal to 3, du= 3dx.
And from that it should be easy to see that dx= (1/3)du.

Quote:

After this step i dont know what to do , do i divide by 3 to produce 1?
Please answer in step .Thank you

I have no idea what you could mean by "do i divide by 3 to produce 1?". Divide what by 3? "Produce 1" where and how? The more precise you are in mathematics, the bettter. Practise being precise.
January 9th 2010, 06:43 AM
VonNemo19

Quote:

Originally Posted by mj.alawami

Question:
$\int \sqrt[3] {3x+5}.dx$

So the equation will be $\int (3x+5)^\frac{1}{3}.dx$

$u=3x+5 \rightarrow \int u^\frac{1}{3}$
$du=3$
After this step i dont know what to do , do i divide by 3 to produce 1?
Please answer in step .Thank you

$u=3x+5$ then

$du=3dx$

$\frac{1}{3}du=dx$ .

So, we have

$\int(3x+5)^{1/3}dx=\int{u^{1/3}}\frac{1}{3}du$
January 9th 2010, 06:46 AM
mj.alawami

Quote:

Originally Posted by HallsofIvy

Here's your problem! du is NOT equal to 3, du= 3dx.
And from that it should be easy to see that dx= (1/3)du.

I have no idea what you could mean by "do i divide by 3 to produce 1?". Divide what by 3? "Produce 1" where and how? The more precise you are in mathematics, the bettter. Practise being precise.

Answer:
$\int \frac{1}{4} (3x+5)^\frac{4}{3} +C$

Is my answer Correct ?
January 9th 2010, 06:49 AM
Prove It

Quote:

Originally Posted by mj.alawami

Question:
$\int \sqrt[3] {3x+5}.dx$

So the equation will be $\int (3x+5)^\frac{1}{3}.dx$

$u=3x+5 \rightarrow \int u^\frac{1}{3}$
$du=3$
After this step i dont know what to do , do i divide by 3 to produce 1?
Please answer in step .Thank you

It might help to write it out like this

$u = 3x + 5$ so $\frac{du}{dx} = 3$ .

Now work on the integral

$\int{(3x + 5)^{\frac{1}{3}}\,dx} = \frac{1}{3}\int{(3x + 5)^{\frac{1}{3}} \cdot 3\,dx}$

$= \frac{1}{3}\int{u^{\frac{1}{3}}\,\frac{du}{dx}\,dx }$

$= \frac{1}{3}\int{u^{\frac{1}{3}}\,du}$ .

Might be an extra step but easier to avoid making mistakes...
January 9th 2010, 06:51 AM
mj.alawami

Quote:

Originally Posted by Prove It

It might help to write it out like this

$u = 3x + 5$ so $\frac{du}{dx} = 3$ .

Now work on the integral

$\int{(3x + 5)^{\frac{1}{3}}\,dx} = \frac{1}{3}\int{(3x + 5)^{\frac{1}{3}} \cdot 3\,dx}$

$= \frac{1}{3}\int{u^{\frac{1}{3}}\,\frac{du}{dx}\,dx }$

$= \frac{1}{3}\int{u^{\frac{1}{3}}\,du}$ .

Might be an extra step but easier to avoid making mistakes...

Answer:
$\frac{1}{2} (3x+5)^\frac{4}{3} +C$

Is my answer Correct ?
January 9th 2010, 06:57 AM
Prove It

No.

$\frac{1}{3}\int{u^{\frac{1}{3}}\,du} = \frac{1}{3}\cdot\frac{3}{4}u^{\frac{4}{3}} + C$

$= \frac{1}{4}u^{\frac{4}{3}} + C$

$= \frac{1}{4}(3x + 5)^{\frac{4}{3}} + C$ .
January 9th 2010, 08:08 AM
HallsofIvy

Quote:

Originally Posted by mj.alawami

Answer:
$\int \frac{1}{4} (3x+5)^\frac{4}{3} +C$

Is my answer Correct ?

What do you get if you differentiate that? Did you even try that?
January 9th 2010, 09:10 PM
bhuang

substitution isn't needed in this question.
If you remember this:

if you have a derivative that is in the form: (ax+b)^n
then the integral will be: [(ax+b)^n+1] / a(n+1)

so just substitute your question into that rule.
January 9th 2010, 09:35 PM
Prove It

Quote:

Originally Posted by bhuang

substitution isn't needed in this question.
If you remember this:

if you have a derivative that is in the form: (ax+b)^n
then the integral will be: [(ax+b)^n+1] / a(n+1)

so just substitute your question into that rule.

Don't try remembering lots of rules, it gets to be too much of a headache.

It's easier to remember a few basic rules and to use substitution for the rest.

Besides, the rule you speak of comes from substitution anyway.