Find \( \int \sqrt{x^2 + 1} \, dx \), - HSC - SSCE Mathematics Advanced - Question 17 - 2023 - Paper 1

The integral ( \int \sqrt{x^2 + 1} , dx ) can be recognized to be related to the form ( k \int f'(x)f(x)^{n} , dx ).

To solve it, let's express ( \sqrt{x^2 + 1} ) in a suitable form:

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

Next, we can apply substitution. Let ( u = x^2 + 1 ), then ( du = 2x , dx ). Thus, we rewrite ( dx ) as ( \frac{du}{2x} ).

Now we have:

$$= \int \frac{x}{2u^{1/2}} \frac{du}{2x} = \frac{1}{2}\int u^{-1/2} du.$$

Integrating gives:

$$= \frac{1}{2} \cdot 2u^{1/2} + C = \sqrt{x^2 + 1} + C.$$

Thus,

$$\int \sqrt{x^2 + 1} \, dx = \sqrt{x^2 + 1} + C.$$