Evaluate the integral: I = integral dx/(3x+ 8th root

Evaluate the integral: $$ I = \int \frac{dx}{3x+\sqrt[8]{x^2}}. $$

Explanation of Method

Since $\sqrt[8]{x^2}=x^{1/4}$, factoring $x^{1/4}$ out of the denominator and substituting $u=x^{1/4}$ reduces the integral to a simple $\int dv/v$ after a second substitution.

Simplify $\sqrt[8]{x^2}=x^{1/4}$.
Substitute $u=x^{1/4}$.
Substitute again $v=3u^3+1$.
Integrate and back-substitute.

Step 1 · Simplify and substitute

$$ 3x+\sqrt[8]{x^2} = x^{1/4}\left(3x^{3/4}+1\right). $$

Let $u=x^{1/4}$, so $x=u^4$, $dx=4u^3\,du$, $x^{3/4}=u^3$:

$$ I = \int\frac{4u^3\,du}{u(3u^3+1)} = 4\int\frac{u^2}{3u^3+1}\,du. $$

Step 2 · Second substitution

Let $v=3u^3+1$, so $dv=9u^2\,du$:

$$ 4\int\frac{u^2\,du}{3u^3+1} = \frac49\int\frac{dv}v = \frac49\ln|v|+C. $$

Step 3 · Back-substitute

$$ I = \frac49\ln\left|3x^{3/4}+1\right| + C. $$

Summary Table

Substitution	Relation
$u=x^{1/4}$	$dx=4u^3\,du$
$v=3u^3+1$	$dv=9u^2\,du$

Additional Notes

The domain requires $x>0$ so that $x^{1/4}$ is real and the denominator is defined.
The result was verified by differentiation: $\dfrac{d}{dx}\left[\tfrac49\ln|3x^{3/4}+1|\right] = \tfrac49\cdot\dfrac{(9/4)x^{-1/4}}{3x^{3/4}+1} = \dfrac{x^{-1/4}}{3x^{3/4}+1} = \dfrac1{3x+x^{1/4}}$, matching the integrand.

Result

The solution to the integral is:

$$ I = \frac49\ln\left|3\sqrt[4]{x^3}+1\right| + C. $$