Integrating Complex Functions Using Integration by Parts

When dealing with complex functions, integration by parts often provides a straightforward approach. This article will demonstrate how to integrate the function (frac{x^2}{(m^2x^2 1)^2}) using this technique, showcasing its application step-by-step.

Step-by-Step Integration Using Integration by Parts

Given the integral:

( mathcal{I}(x) int frac{x^2}{(m^2x^2 1)^2} , dx )

Let's start by making a substitution to simplify the integral.

Set (u mx). Then, (du m , dx) and (dx frac{du}{m}).

Substitute these into the integral:

( mathcal{I}(x) frac{1}{m^3} int frac{left(frac{u}{m}right)^2}{(u^2 1)^2} , frac{du}{m} )

( frac{1}{m^3} int frac{u^2}{m^3 (u^2 1)^2} , du )

( frac{1}{m^3} cdot frac{1}{m^2} int frac{u^2}{(u^2 1)^2} , du )

( frac{1}{m^5} int frac{u^2}{(u^2 1)^2} , du )

Applying Integration by Parts

Now, we use integration by parts. Let:

(u u)

(dv frac{1}{(u^2 1)^2} du)

Then:

(du du)

(v -frac{1}{2(u^2 1)})

The integration by parts formula is:

(uv - int v , du)

Therefore:

( mathcal{I}(x) frac{1}{m^5} left[ -frac{u}{2(u^2 1)} - int -frac{1}{2(u^2 1)} cdot du right] C )

( frac{1}{m^5} left[ -frac{u}{2(u^2 1)} frac{1}{2} int frac{1}{u^2 1} , du right] C )

( frac{1}{m^5} left[ -frac{u}{2(u^2 1)} frac{1}{2} arctan(u) right] C )

Substitute back (u mx):

( mathcal{I}(x) frac{1}{m^5} left[ -frac{mx}{2(m^2x^2 1)} frac{1}{2} arctan(mx) right] C )

Final Integral Expression

Therefore, the integral is:

( mathcal{I}(x) -frac{1}{2m^3} left[ frac{mx}{(m^2x^2 1)} - arctan(mx) right] C_m )

where (C_m -C / 2m^3).

Case Example: m (sqrt{2})

For (m sqrt{2}), the integral simplifies to:

( mathcal{I}(x) -frac{1}{2(sqrt{2})^3} left[ frac{sqrt{2}x}{(2x^2 1)} - arctan(sqrt{2}x) right] C_m )

( -frac{1}{4sqrt{2}} left[ frac{sqrt{2}x}{(2x^2 1)} - arctan(sqrt{2}x) right] C_m )

Conclusion

This approach shows how integration by parts can be effectively used to solve complex integrals involving rational functions. Understanding and practicing these techniques is crucial for anyone delving into advanced calculus and related fields.