Evaluate the integral: I = integral ln (x+ sqrt(1+x²)) dx
Evaluate the integral: $$ I = \int \ln\left(x+\sqrt{1+x^2}\right) dx. $$
Explanation of Method
The expression $\ln(x+\sqrt{1+x^2})$ is exactly the inverse hyperbolic sine function $\sinh^{-1}x$. Recognizing this reduces the problem to integrating $\sinh^{-1}x$ by parts, leaving a simple substitution integral.
- Recognize the integrand as $\sinh^{-1}x$.
- Apply integration by parts with $u=\sinh^{-1}x$, $dv=dx$.
- Evaluate the leftover integral via substitution.
Step 1 · Identify the integrand
Using the hyperbolic substitution $x=\sinh t$, so $\sqrt{1+x^2}=\cosh t$, and the identity $\sinh t+\cosh t=e^t$:
$$ \ln\left(x+\sqrt{1+x^2}\right) = \ln\left(\sinh t+\cosh t\right) = \ln(e^t) = t = \sinh^{-1}x. $$
So the integral becomes:
$$ I = \int \sinh^{-1}x\,dx. $$
Step 2 · Integration by parts
Let:
$$ u = \sinh^{-1}x \quad \Rightarrow \quad du = \frac{dx}{\sqrt{1+x^2}}, \qquad dv = dx \quad \Rightarrow \quad v = x. $$
Applying the formula:
$$ I = x\sinh^{-1}x - \int \frac{x}{\sqrt{1+x^2}}\,dx. $$
Step 3 · Evaluate the leftover integral
Let $w=1+x^2$, so $dw=2x\,dx$:
$$ \int \frac{x}{\sqrt{1+x^2}}\,dx = \int \frac{dw}{2\sqrt{w}} = \sqrt{w} = \sqrt{1+x^2}. $$
Step 4 · Combine
$$ I = x\sinh^{-1}x - \sqrt{1+x^2} + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Integrand identity | $\ln(x+\sqrt{1+x^2}) = \sinh^{-1}x$ |
| $u$ | $\sinh^{-1}x$ |
| $dv$ | $dx$ |
Additional Notes
- The integrand is defined for all real $x$.
- Recognizing $\ln(x+\sqrt{1+x^2})$ as $\sinh^{-1}x$ up front avoids working directly with the awkward composite logarithm-of-a-radical expression; once identified, the problem reduces to the same pattern as $\int\arcsin x\,dx$ or $\int\arctan x\,dx$.
- The result can be checked by differentiating $x\sinh^{-1}x-\sqrt{1+x^2}$ and confirming it returns $\sinh^{-1}x=\ln(x+\sqrt{1+x^2})$.
Result
The solution to the integral is: