1. ## Integrate x^2(x^3+1)^4 dx

Each time try solve this, the answer i get doesn't match the answer in the back of the book. These are the steps i take to solve this integration: (Please correct me where i'm going wrong!)

$\displaystyle \int x^2(x^3+1)^4 dx$

let $\displaystyle u = x^3 + 1$

$\displaystyle \frac{du}{dx} = 3x^2$

$\displaystyle du = 3x^2 dx$

$\displaystyle dx = \frac{1}{3x^2} du$

$\displaystyle \int x^2u^4 dx$

$\displaystyle \int x^2u^4 \frac{1}{3x^2} du$

$\displaystyle \frac{x^3}{3} . \frac{u^5}{5} + c$

$\displaystyle \frac{x^3(u^5)}{15} + c$

$\displaystyle \frac{x^3(x^3 + 1)^5}{15} + c$

The answer in the back of the book is: $\displaystyle \frac{1}{15}(x^3 + 1)^5 + c$

This isn't urgent, however I would love to see where I'm going wrong here. Thanks

2. Woops!!! i just saw where i was going wrong! Sorry if i wasted your time!!!

Just for reference here's how to do it correctly

$\displaystyle \int x^2(x^3+1)^4 dx$

let $\displaystyle u = x^3 + 1$

$\displaystyle \frac{du}{dx} = 3x^2$

$\displaystyle du = 3x^2 dx$

$\displaystyle dx = \frac{1}{3x^2} du$

$\displaystyle \int x^2u^4 dx$

$\displaystyle \int x^2u^4 \frac{1}{3x^2} du$

$\displaystyle \frac{x^2}{3x^2}\int u^4 du$

$\displaystyle \frac{1}{3}\int u^4 du$ (cancel the $\displaystyle x^2$)

$\displaystyle \frac{(u^5)}{15} + c$

$\displaystyle \frac{(x^3 + 1)^5}{15} + c$

The answer in the back of the book is: $\displaystyle \frac{1}{15}(x^3 + 1)^5 + c$

Much better!!

3. Well done for working it out!
Originally Posted by ezzab69
$\displaystyle \frac{x^2}{3x^2}\int u^4 du$
Technically, since x is a variable, not a constant, you can't take it out of the integrand.
In this case x²/x² = 1, but I would still suggest you do the cancellation in the integrand.

4. Thanks for the tip!

5. Originally Posted by ezzab69
Woops!!! i just saw where i was going wrong! Sorry if i wasted your time!!!

Just for reference here's how to do it correctly

$\displaystyle \int x^2(x^3+1)^4 dx$

let $\displaystyle u = x^3 + 1$

$\displaystyle \frac{du}{dx} = 3x^2$

$\displaystyle du = 3x^2 dx$

$\displaystyle dx = \frac{1}{3x^2} du$

$\displaystyle \int x^2u^4 dx$

$\displaystyle \int x^2u^4 \frac{1}{3x^2} du$

$\displaystyle \frac{x^2}{3x^2}\int u^4 du$
Yes, well done. However, I would strongly urge you NOT to write things like this. Since you defined u as a function of x, x is a function of u and you cannot casually take it outside the integral like that! Here, it works because you can cancel the two $\displaystyle x^2$ terms inside the integral.

However, if the problem were, say, $\displaystyle \int x(x^3+1)^4 dx$ and you made the same substitution, you would get $\displaystyle \int x u^4\frac{du}{3x^2}$, that would NOT be the same as $\displaystyle \frac{1}{3x}\int u^4 du$ because you cannot take the "x" outside the integral like that. In fact, this problem, with "$\displaystyle x$" rather than "$\displaystyle x^2$" cannot be integrated with a substitution- you would have to multiply out the $\displaystyle (x^3+ 1)^4$.

$\displaystyle \frac{1}{3}\int u^4 du$ (cancel the $\displaystyle x^2$)

$\displaystyle \frac{(u^5)}{15} + c$

$\displaystyle \frac{(x^3 + 1)^5}{15} + c$

The answer in the back of the book is: $\displaystyle \frac{1}{15}(x^3 + 1)^5 + c$

Much better!!
I see that The Coffee Machine said the same thing, more succinctly.

6. Yep so as i got further through the book i came across the situation you were talking about. Thanks for clearing things up for me. This forum is great, i should come here more often. Cheers guys!