Evaluate the integral (for x>0): I = integral dx/(x
Evaluate the integral (for $x>0$): $$ I = \int \frac{dx}{x\sqrt{1+x^6}}. $$
Explanation of Method
The substitution $u=x^3$ turns the integral into the standard form $\tfrac13\int du/[u\sqrt{1+u^2}]$, resolved via a hyperbolic ($u=\sinh t$) or direct reciprocal-type substitution.
- Substitute $u=x^3$.
- Apply the standard formula for $\int du/[u\sqrt{1+u^2}]$.
- Back-substitute.
Step 1 · Substitute
Let $u=x^3$, so $du=3x^2\,dx$, and $x^6=u^2$:
$$ I = \int\frac{dx}{x\sqrt{1+x^6}} = \int\frac{x^2\,dx}{x^3\sqrt{1+x^6}} = \frac13\int\frac{du}{u\sqrt{1+u^2}}. $$
Step 2 · Apply the standard formula
Using $u=\sinh t$, $\sqrt{1+u^2}=\cosh t$, this reduces to $\int\text{csch}(t)\,dt=\ln\left|\tanh(t/2)\right|+C$, which in terms of $u$ gives the standard result:
$$ \int\frac{du}{u\sqrt{1+u^2}} = \ln\left|\frac u{1+\sqrt{1+u^2}}\right| + C. $$
Step 3 · Combine and back-substitute
$$ I = \frac13\ln\left|\frac u{1+\sqrt{1+u^2}}\right| + C. $$
With $u=x^3$, and splitting the logarithm:
Summary Table
| Substitution | Effect |
|---|---|
| $u=x^3$ | reduces to $\tfrac13\int du/[u\sqrt{1+u^2}]$ |
Additional Notes
- The domain is $x>0$ (a symmetric derivation handles $x<0$).
- The result was verified by differentiation at $x=1$ and $x=2$: the derivative matches the integrand $1/[x\sqrt{1+x^6}]$ exactly at both points.
Result
The solution to the integral is: