Question

Verify by differentiation that the formula is correct.

\(\int {\frac{x}{{\sqrt {{x^2} + 1} }}} dx = \sqrt {{x^2} + 1} + C\)

Short answer

The given Formula is verified.

Step-by-step solution

Find the antiderivative.

\(\int {f(x)dx = F(x)} \)means that\({F^'}(x) = f(x)\)

Here,\(\int {f(x)dx} \)is antiderivative of\(\int {f(x)} \)

Calculate the antiderivative of the given function

Take R.H.S and differentiate with respect to x.

\(\begin{aligned}{c}\frac{d}{{dx}}\left( {\sqrt {{x^2} + 1} + C} \right) = \frac{d}{{dx}}\sqrt {{x^2} + 1} + \frac{d}{{dx}}(C)\\ = \frac{1}{{2\sqrt {{x^2} + 1} }} \times 2x + 0 = \frac{x}{{\sqrt {{x^2} + 1} }}\end{aligned}\)

Hence, the antiderivative of the function\(\frac{x}{{\sqrt {{x^2} + 1} }}\)is \(\sqrt {{x^2} + 1} + C\)

Therefore, the formula \(\int {\frac{x}{{\sqrt {{x^2} + 1} }}} dx = \sqrt {{x^2} + 1} + C\) is verified.