When the integral of products is the product of integrals.

Question

I'm self-studying and was doing the following integral:

$$I = \int \frac{e^{\frac{1}{x}+\tan^{-1}x}}{x^2+x^4} dx $$

I solved it fine by letting $ u = \frac{1}{x} + \tan^{-1}x$.

My question is about an alternative method I saw in which it seems the product rule was not applied:

$$ I = \int \left(\frac { e^{\frac{1}{x}}} {x^2}\right) \left( \frac{e^{\tan^{-1}x}}{x^2+1}\right) dx $$

$$ = \int \frac {e^{\frac{1}{x}}}{x^2} dx \cdot \int \frac{e^{\tan^{-1}x}}{x^2+1}dx$$

Completing the work following this step leads to the same solution as I originally found.

It is this step that has confused me. I have checked using Wolfram and the two statements are equivalent but I do not understand why.

Why are we able to write the integral of products as the product of integrals here, and not apply the product rule?

Thanks in advance.

Answer 1

Why are we able to write the integral of products as the product of integrals here?

Assume you have two differentiable functions $f,g$ such that $$ f'+g'=f'\cdot g' \tag1 $$ by multiplying by $\displaystyle e^{f+g}$ one gets $$ (f'+g')\cdot e^{f+g}=\left(f'e^{f} \right)\cdot \left(g'e^{g} \right) \tag2 $$ then by integrating both sides $$ e^{f+g}=\int\left(f'e^{f} \right)\cdot \left(g'e^{g} \right) \tag3 $$ since $\displaystyle e^f=\int\left(f'e^{f} \right) $ and $\displaystyle e^g=\int\left(g'e^{g} \right)$ we have

$$ \int\left(f'e^{f} \right)\cdot \int\left(g'e^{g} \right) =\int\left(f'e^{f} \right)\cdot \left(g'e^{g} \right). \tag4 $$

By taking, $f'=-\dfrac1{x^2}$ and $g'=\dfrac1{1+x^2}$ we have $$ f'+g'=-\frac1{x^2}+\frac1{1+x^2}=-\frac1{x^2(1+x^2)}=f'g' $$ which leads to $(4)$ with the given example.

Answer 2

Let $F(x)$, $G(x)$ and $H(x)$ be antiderivatives of $f(x)$, $g(x)$ and $f(x) g(x)$ respectively. If $F(x) G(x) = H(x)$, then differentiating that equation gives us

$$ f(x) G(x) + F(x) g(x) = f(x) g(x) $$

or

$$ f(x) + F(x) \frac{g(x)}{G(x) - g(x)} = 0 $$

(assuming $G(x) \ne g(x)$). Given differentiable $G(x)$, with $g(x) = G'(x)$ and assuming $G(x) \ne g(x)$, you could get a suitable function $F(x)$ by solving the differential equation

$$ y'(x) + y(x) \frac{g(x)}{G(x) - g(x)} = 0$$

EDIT: In the case at hand we may take $g(x) = e^{\arctan(x)}/(x^2+1)$ and $G(x) = e^{\arctan(x)}$. The differential equation simplifies to $$ x^2 y'(x) + y(x) = 0 $$ which has the solutions $$ y(x) = C e^{1/x}$$ and (for $C=1$) this is your $F(x)$.