Evaluate the integral: I = integral sqrt((ln (x+
Evaluate the integral: $$ I = \int \sqrt{\frac{\ln \left(x+\sqrt{1+x^2}\right)}{1+x^2}} \, dx. $$
Explanation of Method
The expression $\ln\left(x+\sqrt{1+x^2}\right)$ is the inverse hyperbolic sine of $x$, whose derivative is the standard $\frac{1}{\sqrt{1+x^2}}$. Substituting $u$ for this logarithm converts the entire integrand — including the $\sqrt{1+x^2}$ that appears implicitly — into the simple form $\sqrt u\,du$.
- Substitute $u=\ln\left(x+\sqrt{1+x^2}\right)$.
- Simplify $du$ using the standard derivative of the inverse hyperbolic sine.
- Rewrite the integral entirely in terms of $u$.
- Integrate and substitute back.
Step 1 · Substitution
We use the substitution:
$$ u = \ln\left(x+\sqrt{1+x^2}\right). $$
Differentiating both sides:
$$ du = \frac{1}{x+\sqrt{1+x^2}}\left(1+\frac{x}{\sqrt{1+x^2}}\right) dx. $$
Since:
$$ 1+\frac{x}{\sqrt{1+x^2}} = \frac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}}, $$
we obtain:
$$ du = \frac{dx}{\sqrt{1+x^2}}. $$
Step 2 · Rewrite the integral
Thus, the given integral transforms into:
$$ I = \int \sqrt{\frac{u}{1+x^2}}\cdot \sqrt{1+x^2}\,du. $$
Simplifying:
$$ I = \int \sqrt{u}\,du. $$
Step 3 · Integrate
Using the standard formula:
$$ \int u^{1/2}\,du = \frac{2}{3}u^{3/2}, $$
we obtain:
$$ I = \frac{2}{3}u^{3/2} + C. $$
Step 4 · Substitute back
Substituting back $u = \ln\left(x+\sqrt{1+x^2}\right)$:
$$ I = \frac{2}{3}\left(\ln\left(x+\sqrt{1+x^2}\right)\right)^{3/2} + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $u = \ln\left(x+\sqrt{1+x^2}\right)$ |
| $du$ | $dx/\sqrt{1+x^2}$ |
| Antiderivative | $\tfrac23 u^{3/2}$ |
Additional Notes
- The domain of $I$ requires $\ln\left(x+\sqrt{1+x^2}\right) \geq 0$, i.e. $x\geq 0$, since $x+\sqrt{1+x^2}\geq 1$ only when $x\geq 0$ (this expression equals $\operatorname{arsinh}(x)$, which is negative for $x<0$, making the square root of it undefined in the reals).
- Recognizing $\ln\left(x+\sqrt{1+x^2}\right)$ as $\operatorname{arsinh}(x)$ and recalling its standard derivative $\dfrac{1}{\sqrt{1+x^2}}$ is what makes the $\sqrt{1+x^2}$ factors cancel so cleanly between the substitution and the original integrand.
- The result can be checked by differentiating $\tfrac23\left(\ln(x+\sqrt{1+x^2})\right)^{3/2}$ with the chain rule and confirming it returns the original integrand.
Result
The solution to the integral is: