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

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

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

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

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

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

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

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.

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

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.

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

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.

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

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

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