Evaluate the integral: I = integral (x dx)/((x²+1)
Evaluate the integral: $$ I = \int \frac{x\,dx}{(x^2+1) \sqrt{x^2+1}}. $$
Explanation of Method
Since $(x^2+1)\sqrt{x^2+1} = (x^2+1)^{3/2}$, the integrand is really $x(x^2+1)^{-3/2}$. Substituting $u=x^2+1$ converts this into a routine negative-fractional-power integral in $u$.
- Rewrite the denominator as a single power of $x^2+1$.
- Substitute $u$ for $x^2+1$ and match $x\,dx$ to $du$.
- Integrate using the power rule.
- Substitute back and add the constant of integration.
Step 1 · Substituting $u = x^2 + 1$
To solve this integral, we use the substitution method. Observe that the denominator contains $x^2+1$ and its square root. Let:
$$ u = x^2 + 1. $$
Differentiating both sides:
$$ du = 2x\,dx. $$
Rewriting:
$$ \frac{du}{2} = x\,dx. $$
Now, substituting into the integral:
$$ I = \int \frac{\frac{du}{2}}{u\sqrt{u}}. $$
Step 2 · Simplifying the integral
Rewriting the denominator:
$$ I = \frac{1}{2}\int \frac{du}{u^{3/2}}. $$
Using the power rule:
$$ \int u^{-3/2}\,du = \frac{u^{-1/2}}{-1/2} = -2u^{-1/2}. $$
Step 3 · Evaluating and substituting back
$$ I = \frac{1}{2}\times(-2)u^{-1/2} + C = -u^{-1/2} + C. $$
Since $u = x^2+1$, we substitute back:
$$ I = -\frac{1}{\sqrt{x^2+1}} + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Rewritten integrand | $x(x^2+1)^{-3/2}$ |
| Substitution | $u = x^2+1$ |
| Antiderivative | $-u^{-1/2}$ |
Additional Notes
- The integrand is defined for all real $x$, since $x^2+1 \geq 1 > 0$ always.
- Recognizing $(x^2+1)\sqrt{x^2+1}$ as $(x^2+1)^{3/2}$ before substituting simplifies the bookkeeping considerably compared to treating the product and root as separate factors.
- The result can be checked by differentiating $-\dfrac{1}{\sqrt{x^2+1}}$ with the chain rule and confirming it returns $\dfrac{x}{(x^2+1)\sqrt{x^2+1}}$.
Result
The solution to the integral is: