integral of 1/(12sqrt(x)+12x) — step-by-step solution

1) Before integrating, let's calculate:

\frac{1}{12 \sqrt{x} + 12 x}

ⓘ What the terms have in common is factored out, so that the fraction can be simplified:

\frac{1}{12 \sqrt{x} + 12 x}

\frac{1}{12 \sqrt{x} \cdot (1 + \sqrt{x})}

\frac{1}{12 \sqrt{x} \cdot (1 + \sqrt{x})}

2) Let's compute the integral:

\int \frac{1}{12 \sqrt{x} \cdot (1 + \sqrt{x})} d x

For the math expression:

\frac{1}{12 \sqrt{x} \cdot (1 + \sqrt{x})}

, since (ln(a))' = a'/a, we try differentiating

\ln (1 + \sqrt{x})

3) Let's perform the differentiation:

\frac{d}{d x} (\ln (1 + \sqrt{x}))

We differentiate using:

\frac{d}{d x} (\ln (f)) = \frac{d}{d x} (f) \cdot \frac{1}{f}

(where f =

1 + \sqrt{x}

)

ⓘ The derivative of ln(x) is 1/x:

\frac{d}{d x} (\ln (1 + \sqrt{x}))

\frac{d}{d x} (1 + \sqrt{x}) \cdot \frac{1}{1 + \sqrt{x}}

\frac{d}{d x} (1 + \sqrt{x}) \cdot \frac{1}{1 + \sqrt{x}}

ⓘ Multiplying a fraction by a term multiplies its numerator:

\frac{d}{d x} (1 + \sqrt{x}) \cdot \frac{1}{1 + \sqrt{x}}

\frac{\frac{d}{d x} (1 + \sqrt{x}) \cdot 1}{1 + \sqrt{x}}

\frac{\frac{d}{d x} (1 + \sqrt{x})}{1 + \sqrt{x}}

■The derivative of a sum is the sum of the derivatives:

\frac{d}{d x} (1 + \sqrt{x})

0 + \frac{d}{d x} (\sqrt{x})

\frac{0 + \frac{d}{d x} (\sqrt{x})}{1 + \sqrt{x}}

ⓘ Adding 0 changes nothing:

0 + \frac{d}{d x} (\sqrt{x})

\frac{d}{d x} (\sqrt{x})

■The derivative of sqrt(x) is 1/(2*sqrt(x)):

\frac{d}{d x} (\sqrt{x})

\frac{1}{2 \sqrt{x}}

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

ⓘ Dividing a fraction by

1 + \sqrt{x}

is multiplying its denominator by

1 + \sqrt{x}

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

\frac{1}{2 \sqrt{x} \cdot (1 + \sqrt{x})}

\frac{1}{2 \sqrt{x} \cdot (1 + \sqrt{x})}

You observe that the computed derivative and the original math expression are the same, except for a coefficient. So the integral is:

ⓘ The derivative of

\frac{1}{6} \cdot \ln (1 + \sqrt{x})

\frac{1}{12 \sqrt{x} \cdot (1 + \sqrt{x})}

, so

\frac{1}{6} \cdot \ln (1 + \sqrt{x})

is an antiderivative of

\frac{1}{12 \sqrt{x} \cdot (1 + \sqrt{x})}

\int \frac{1}{12 \sqrt{x} \cdot (1 + \sqrt{x})} d x

\frac{1}{6} \cdot \ln (1 + \sqrt{x})

\frac{1}{6} \cdot \ln (1 + \sqrt{x})

ⓘ Multiplying a fraction by a term multiplies its numerator:

\frac{1}{6} \cdot \ln (1 + \sqrt{x})

\frac{1 \cdot \ln (1 + \sqrt{x})}{6}

\frac{\ln (1 + \sqrt{x})}{6}

\frac{\ln (1 + \sqrt{x})}{6}

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

\frac{\ln (\sqrt{x} + 1)}{6} + C