Evaluate the integral: integral (x³ dx)/(x^8 - 2)

Evaluate the integral: $$ \int \frac{x^3\,dx}{x^8 - 2}. $$

Step 1 · Substitution

We will solve this integral using substitution. Let:

$$ u = x^4. $$

Then, differentiate both sides with respect to $x$:

$$ \frac{du}{dx} = 4x^3, \quad \text{so} \quad du = 4x^3\,dx. $$

Thus, $x^3\,dx = \dfrac{du}{4}$.

Step 2 · Substitute into the integral

Now, substitute into the original integral:

$$ \int \frac{x^3\,dx}{x^8-2} = \int \frac{\frac{du}{4}}{u^2-2} = \frac{1}{4}\int \frac{du}{u^2-2}. $$

Step 3 · Apply the standard result

Next, we recognize that the integral $\displaystyle \int \frac{du}{u^2-a^2}$ is a standard form, which can be solved as:

$$ \frac{1}{2a}\ln\left|\frac{u-a}{u+a}\right|. $$

In our case, $a=\sqrt2$, so we have:

$$ \frac{1}{4}\int \frac{du}{u^2-2} = \frac{1}{4}\cdot \frac{1}{2\sqrt2}\ln\left|\frac{u-\sqrt2}{u+\sqrt2}\right| = \frac{1}{8\sqrt2}\ln\left|\frac{u-\sqrt2}{u+\sqrt2}\right|. $$

Step 4 · Substitute back

Now, substitute $u=x^4$ back into the result:

$$ \frac{1}{8\sqrt{2}}\ln\left|\frac{x^4-\sqrt{2}}{x^4+\sqrt{2}}\right| + C. $$

Summary Table

Additional Notes

• The domain of $I$ requires $x^4 \neq \sqrt2$, i.e. $x \neq \pm 2^{1/8}$, where the denominator $u^2-2$ vanishes.
• Recognizing $x^8$ as $(x^4)^2$ is the key first step here, mirroring the same idea used in Problem 73 for $x^4=(x^2)^2$.
• The result can be checked by differentiating $\tfrac{1}{8\sqrt2}\ln\left|\tfrac{x^4-\sqrt2}{x^4+\sqrt2}\right|$ with the chain rule and confirming it returns $\dfrac{x^3}{x^8-2}$.