integral of 1/(sqrt(64+x^2)) — step-by-step solution

Let's compute the integral:

\int \frac{1}{\sqrt{64 + x^{2}}} d x

The derivative of

\ln (x + \sqrt{64 + x^{2}})

\frac{1 + \frac{x}{\sqrt{64 + x^{2}}}}{x + \sqrt{64 + x^{2}}}

\frac{\frac{\sqrt{64 + x^{2}} + x}{\sqrt{64 + x^{2}}}}{x + \sqrt{64 + x^{2}}}

\frac{1}{\sqrt{64 + x^{2}}}

ⓘ The derivative of

\ln (x + \sqrt{64 + x^{2}})

\frac{1}{\sqrt{64 + x^{2}}}

, so

\ln (x + \sqrt{64 + x^{2}})

is an antiderivative of

\frac{1}{\sqrt{64 + x^{2}}}

\int \frac{1}{\sqrt{64 + x^{2}}} d x

\ln (x + \sqrt{64 + x^{2}})

\ln (x + \sqrt{64 + x^{2}})

▶Indefinite integrals are defined up to an additive constant C, so this yields:

\ln (x + \sqrt{x^{2} + 64}) + C