Q10:

Choose the correct answer in Exercises 10 to 11. 10. $\int \sqrt{1+x^2} dx$ is equal to

Step 1: Establishing the Standard Integration Formula

To evaluate the integral $I = \int \sqrt{1+x^2} \, dx$, we utilize the standard algebraic form for the integral of the square root of a quadratic expression, specifically $\int \sqrt{x^2 + a^2} \, dx$.

The standard integration formula derived via integration by parts is:

$\int \sqrt{x^2 + a^2} \, dx = \frac{x}{2} \sqrt{x^2 + a^2} + \frac{a^2}{2} \ln \left| x + \sqrt{x^2 + a^2} \right| + C$ [By the method of Integration by Parts, where $\int u \, dv = uv - \int v \, du$]

Step 2: Identifying Given Parameters

Given the integral $\int \sqrt{1+x^2} \, dx$, we compare this to the standard form $\int \sqrt{x^2 + a^2} \, dx$:

• $a^2 = 1 \implies a = 1$

Step 3: Substitution and Calculation

Substituting $a = 1$ into the established formula:

$I = \frac{x}{2} \sqrt{x^2 + 1^2} + \frac{1^2}{2} \ln \left| x + \sqrt{x^2 + 1^2} \right| + C$

Simplifying the constants and radicands:

$I = \frac{x}{2} \sqrt{x^2 + 1} + \frac{1}{2} \ln \left| x + \sqrt{x^2 + 1} \right| + C$

Visual Representation of the Integration Function

Step 4: Analytical Verification

To verify, differentiate the resulting expression $f(x) = \frac{x}{2} \sqrt{x^2 + 1} + \frac{1}{2} \ln |x + \sqrt{x^2+1}|$:

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

$= \left( \frac{1}{2} \sqrt{x^2+1} + \frac{x}{2} \cdot \frac{2x}{2\sqrt{x^2+1}} \right) + \frac{1}{2} \left( \frac{1 + \frac{x}{\sqrt{x^2+1}}}{x + \sqrt{x^2+1}} \right)$

$= \frac{x^2+1 + x^2}{2\sqrt{x^2+1}} + \frac{1}{2} \left( \frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}(x+\sqrt{x^2+1})} \right) = \frac{2x^2+1}{2\sqrt{x^2+1}} + \frac{1}{2\sqrt{x^2+1}} = \frac{2x^2+2}{2\sqrt{x^2+1}} = \sqrt{x^2+1}$

[Verification confirms the derivative returns the original integrand]

Final Solution: $\frac{2}{x} \sqrt{x^2+1} + \frac{1}{2} \ln |x + \sqrt{x^2+1}| + C$