Indefinite integration involving exponential and rational function

[juantheron]

Calculation of $\displaystyle \int e^x \cdot \frac{x^3-x+2}{(x^2+1)^2}dx$

 

[Theia]

Spoiler

Let \(\displaystyle y = \mathrm{e}^x \cdot \frac{x^3 - x + 2}{(x^2 + 1)^2} = f'\) for some function \(\displaystyle f\).

If we then write \(\displaystyle f = \mathrm{e}^x q\) for some function \(\displaystyle q\) and differentiate this we see that \(\displaystyle f' = \mathrm{e}^x (q + q')\). Thus we can write

\(\displaystyle \mathrm{e}^x \cdot \frac{x^3 - x + 2}{(x^2 + 1)^2} = \mathrm{e}^x (q + q') \quad \Leftrightarrow \quad \frac{x^3 - x + 2}{(x^2 + 1)^2} = q + q'\).

This could be solved directly, but I liked more to do this light differently like this:

Now I assume that function \(\displaystyle q\) can be written in form \(\displaystyle q = \frac{p}{x^2 + 1}\) using some function \(\displaystyle p\). Substituting this into the equation gives us

\(\displaystyle (x - 1)^2p + (x^2 + 1)p' = x^3 - x + 2\).

This ODE is easy to solve. For homogenous equation we obtain

\(\displaystyle p_h = C\mathrm{e}^{-x}(x^2 + 1)\)

and particular solution

\(\displaystyle p_p = x + 1\),

and thus the solution to the ODE is

\(\displaystyle p = x + 1 + C\mathrm{e}^{-x}(x^2 + 1)\).

Now we can write the functions \(\displaystyle q\) and \(\displaystyle f\), namely

\(\displaystyle q = \frac{p}{x^2 + 1} = \frac{x + 1}{x^2 + 1} + C\mathrm{e}^{-x}\)

and

\(\displaystyle f = \mathrm{e}^x q = \mathrm{e}^x \cdot \frac{x + 1}{x^2 + 1} + C\).

Hence

\(\displaystyle \int \mathrm{e}^x \cdot \frac{x^3 - x + 2}{(x^2 + 1)^2} \mathrm{d}x = \mathrm{e}^x \cdot \frac{x + 1}{x^2 + 1} + C\).