# Thread: Can somebody help me understand why for this indefinite integral?

1. ## Can somebody help me understand why for this indefinite integral?

I am being asked to find the indefinite integral
$\displaystyle \int x(x^{2}-1)^3dx$
The solution reads as:
The factor $\displaystyle x$ in the integrand is, except for a constant multiple, the derivative of $\displaystyle x^{2}-1$,
so that we can apply equation
$\displaystyle \int[f(x)]^{n}f'(x)dx=\frac{1}{n+1}[f(x)]^{n+1}+c$
with $\displaystyle f(x)=x^{2}-1$ and $\displaystyle n=3$.
Since $\displaystyle f'(x)=2x$, we write $\displaystyle x=\frac{1}{2}(2x)$ before applying the formula.
Thus we have
$\displaystyle \int x(x^{2}-1)^{3}dx=\frac{1}{2}\int(x^{2}-1)^{3}(2x)dx$
Why the $\displaystyle (2x)$ and why the $\displaystyle \frac{1}{2}$ outside of the integral?
Many thanks all.

2. ## Re: Can somebody help me understand why for this indefinite integral?

Because $\displaystyle (x^2-1)^3(2x)$ precisely equals $\displaystyle [f(x)]^3f'(x)$ for $\displaystyle f(x)=x^2-1$. We want to represent the integrand in this form because, as the text says,

$\displaystyle \int[f(x)]^{n}f'(x)dx=\frac{1}{n+1}[f(x)]^{n+1}+c$

3. ## Re: Can somebody help me understand why for this indefinite integral?

So if I see a question like this, I should find the derivative of what's inside the brackets, if it matches what's outside the brackets (but has a numerical multiple) then I use the equation above, but include a fraction to counter the numerical multiple so it equals 1? And the numerical fraction (in this case a half to cancel the 2 to a 1) goes outside the integral to comply with the constant multiple rule?

(struggling a little here)

4. ## Re: Can somebody help me understand why for this indefinite integral?

The penny has dropped, I get it now,
many thanks.

5. ## Re: Can somebody help me understand why for this indefinite integral?

Yes. In general, you massage the expression in whatever way to give it a form for which one knows how to find an integral.

6. ## Re: Can somebody help me understand why for this indefinite integral?

Originally Posted by emakarov
Because $\displaystyle (x^2-1)^3(2x)$ precisely equals $\displaystyle [f(x)]^3f'(x)$ for $\displaystyle f(x)=x^2-1$. We want to represent the integrand in this form because, as the text says,

$\displaystyle \int[f(x)]^{n}f'(x)dx=\frac{1}{n+1}[f(x)]^{n+1}+c$
Would I be right in saying this is the substitution method i.e. the above LHS is equivalent to

$\displaystyle \int f(g(x))g'(x)dx$ where $\displaystyle f(g(x))$would be equal to $\displaystyle [f(x)]^{n}$

I get the same answer with this version. Thanks

7. ## Re: Can somebody help me understand why for this indefinite integral?

Yes, this is the substitution method.

8. ## Re: Can somebody help me understand why for this indefinite integral?

Originally Posted by emakarov
Yes, this is the substitution method.
Would I be correct in saying that is the product form of the method. Would the quotient form be the inverse i.e.

$\displaystyle \int [f(x)]^{-n} f'(x) dx$ or $\displaystyle \int [f(x)]^{n} f'(x)^{-1} dx$

Would these work? Thanks

9. ## Re: Can somebody help me understand why for this indefinite integral?

Originally Posted by bugatti79
Would I be correct in saying that is the product form of the method. Would the quotient form be the inverse i.e.

$\displaystyle \int [f(x)]^{-n} f'(x) dx$ or $\displaystyle \int [f(x)]^{n} f'(x)^{-1} dx$

Would these work? Thanks
no ... in both cases of substitution, if u = f(x) , then the "du" is f'(x) dx

10. ## Re: Can somebody help me understand why for this indefinite integral?

Originally Posted by bugatti79
Would the quotient form be the inverse i.e.

$\displaystyle \int [f(x)]^{-n} f'(x) dx$ or $\displaystyle \int [f(x)]^{n} f'(x)^{-1} dx$
Suppose that $\displaystyle G'(y)=g(y)$ for some functions $\displaystyle G(y)$, $\displaystyle g(y)$. Then $\displaystyle \int g(f(x))f'(x)\,dx=\int [G(f(x))]'\,dx=G(f(x))+C$. This works whether $\displaystyle g(y)$ is $\displaystyle y^n$ or $\displaystyle y^{-n}$.

So, $\displaystyle \int [f(x)]^{-n} f'(x) dx=\frac{f(x)^{-n+1}}{-n+1}+C$ for $\displaystyle n\ne 1$. Here $\displaystyle g(y)=y^{-n}$ and $\displaystyle G(y)=\frac{y^{-n+1}}{-n+1}.$

I don't see how to apply the substitution method to $\displaystyle \int [f(x)]^{n} f'(x)^{-1} dx$.