Evaluate the integral: I = integral dx/(x³sqrt(x²+1))
Evaluate the integral: $$ I = \int \frac{dx}{x^3\sqrt{x^2+1}}. $$
Explanation of Method
The reciprocal substitution $u=1/x$ turns the integral into $-\int u\,du/\sqrt{1+u^2}$, a direct power-rule form after writing $\sqrt{x^2+1}=\sqrt{1+u^2}/|u|$.
- Substitute $u=1/x$.
- Integrate using the power rule.
- Back-substitute and add the constant piece from the standard form.
Step 1 · Substitute
Let $u=1/x$, so $x=1/u$, $dx=-du/u^2$, and (for $x>0$) $x^3\sqrt{x^2+1}=\sqrt{1+u^2}/u^4$:
$$ I = \int\frac{-du/u^2}{\sqrt{1+u^2}/u^4} = -\int\frac{u^2\,du}{\sqrt{1+u^2}}. $$
Step 2 · Integrate
Using the standard reduction $\int\dfrac{u^2\,du}{\sqrt{1+u^2}}=\dfrac u2\sqrt{1+u^2}-\dfrac12\ln\left|u+\sqrt{1+u^2}\right|+C$:
$$ I = -\frac u2\sqrt{1+u^2} + \frac12\ln\left|u+\sqrt{1+u^2}\right| + C. $$
Step 3 · Back-substitute
With $u=1/x$ (for $x>0$), $\sqrt{1+u^2}=\sqrt{x^2+1}/x$:
$$ I = -\frac{\sqrt{x^2+1}}{2x^2} + \frac12\ln\left|\frac{1+\sqrt{x^2+1}}x\right| + C. $$
Summary Table
| Piece | Antiderivative |
|---|---|
| $u^2/\sqrt{1+u^2}$ | $\tfrac u2\sqrt{1+u^2}-\tfrac12\ln|u+\sqrt{1+u^2}|$ |
Additional Notes
- The domain excludes $x=0$.
- The result was verified by differentiation at $x=1$: the boxed expression's derivative evaluates to $1/\sqrt2\approx0.7071$, matching the integrand $1/(x^3\sqrt{x^2+1})=1/(1\cdot\sqrt2)\approx0.7071$ exactly.
Result
The solution to the integral is: