Question Statement

Evaluate the integral:

$\int \ln \left[x+\sqrt{1+x^2}\right]dx$

---

Background and Explanation

This integral requires integration by parts, a technique based on the product rule for differentiation. The integrand $\ln \left(x+\sqrt{1+x^2}\right)$ is actually the inverse hyperbolic sine function $\text{arsinh}(x)$, though we will solve it using standard algebraic manipulation and careful differentiation of the composite logarithmic function.

---

Solution

We begin by setting up the integral for integration by parts, treating the logarithmic term as one function and $1$ as the other.

Let: $I=\int \ln \left(x+\sqrt{1+x^2}\right)dx=\int \ln \left(x+\sqrt{1+x^2}\right)\cdot 1dx$

Step 1: Apply Integration by Parts

Using the formula $\int udv=uv-\int vdu$, we choose:

• $u=\ln \left(x+\sqrt{1+x^2}\right)$, so we will need to find $du$
• $dv=1dx$, so $v=x$

This gives us: $I=x\ln \left(x+\sqrt{1+x^2}\right)-\int x\cdot \frac{d}{dx}\left[\ln \left(x+\sqrt{1+x^2}\right)\right]dx$

Step 2: Differentiate the Logarithmic Term

Using the chain rule, we compute the derivative: $\frac{d}{dx}\left[\ln \left(x+\sqrt{1+x^2}\right)\right]=\frac{1}{x+\sqrt{1+x^2}}\cdot \frac{d}{dx}\left(x+\sqrt{1+x^2}\right)$

The derivative inside is: $\frac{d}{dx}\left(x+\sqrt{1+x^2}\right)=1+\frac{1}{2}(1+x^2{)}^{-1/2}\cdot 2x=1+\frac{x}{\sqrt{1+x^2}}$

Therefore: $\frac{d}{dx}\left[\ln \left(x+\sqrt{1+x^2}\right)\right]=\frac{1}{x+\sqrt{1+x^2}}\cdot \left(1+\frac{x}{\sqrt{1+x^2}}\right)$

Step 3: Simplify the Integrand

Substituting back into our integral: $I=x\ln \left(x+\sqrt{1+x^2}\right)-\int x\cdot \frac{1}{x+\sqrt{1+x^2}}\cdot \left(\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}\right)dx$

Notice that the numerator $\sqrt{1+x^2}+x$ cancels with the denominator $x+\sqrt{1+x^2}$: $I=x\ln \left(x+\sqrt{1+x^2}\right)-\int \frac{x}{\sqrt{1+x^2}}dx$

Or equivalently: $I=x\ln \left(x+\sqrt{1+x^2}\right)-\int x(1+x^2{)}^{-1/2}dx$

Step 4: Evaluate the Remaining Integral

To integrate $\int x(1+x^2{)}^{-1/2}dx$, we use the substitution method (or recognize the pattern). We rewrite it as: $\int x(1+x^2{)}^{-1/2}dx=\frac{1}{2}\int (1+x^2{)}^{-1/2}\cdot 2xdx$

Using the power rule for integration $\int u^{n}du=\frac{u^{n+1}}{n+1}$ where $u=1+x^2$ and $n=-\frac{1}{2}$: $=\frac{1}{2}\cdot \frac{(1+x^2{)}^{1/2}}{1/2}=\sqrt{1+x^2}$

Step 5: Final Result

Substituting back: $I=x\ln \left(x+\sqrt{1+x^2}\right)-\sqrt{1+x^2}+c$

---

Key Formulas or Methods Used

• Integration by Parts: $\int udv=uv-\int vdu$
• Chain Rule for Differentiation: $\frac{d}{dx}[\ln (u)]=\frac{1}{u}\cdot \frac{du}{dx}$
• Derivative of Square Root: $\frac{d}{dx}\sqrt{1+x^2}=\frac{x}{\sqrt{1+x^2}}$
• Power Rule for Integration: $\int u^{n}du=\frac{u^{n+1}}{n+1}+c$ (for $n\ne -1$)
• Algebraic Simplification: Recognizing that $\frac{x+\sqrt{1+x^2}}{x+\sqrt{1+x^2}}=1$ to cancel terms

---

Summary of Steps

1. Set up integration by parts with $u=\ln (x+\sqrt{1+x^2})$ and $dv=dx$
2. Differentiate the logarithmic term using the chain rule, obtaining $\frac{1}{x+\sqrt{1+x^2}}\cdot \left(1+\frac{x}{\sqrt{1+x^2}}\right)$
3. Simplify the resulting integrand by combining fractions and canceling the common factor $(x+\sqrt{1+x^2})$ to get $\frac{x}{\sqrt{1+x^2}}$
4. Integrate $\int \frac{x}{\sqrt{1+x^2}}dx$ using substitution (or recognition) to obtain $\sqrt{1+x^2}$
5. Combine terms to get the final answer: $x\ln (x+\sqrt{1+x^2})-\sqrt{1+x^2}+c$