integral of (ln(y^2))/(3y) — step-by-step solution

1) Let's compute the integral:

\int \frac{\ln (y^{2})}{3 y} d y

For the math expression:

\frac{\ln (y^{2})}{3 y}

, since (a^2)' = 2*a'*a, we try differentiating

{\ln (y^{2})}^{2}

2) Let's perform the differentiation:

\frac{d}{d y} ({\ln (y^{2})}^{2})

Apply the differentiation rule:

\frac{d}{d y} (f^{n}) = n \cdot \frac{d}{d y} (f) \cdot f^{n - 1}

with f =

\ln (y^{2})

ⓘ The derivative of x^n is n*x^(n-1):

\frac{d}{d y} ({\ln (y^{2})}^{2})

2 \cdot \frac{d}{d y} (\ln (y^{2})) \cdot \ln (y^{2})

2 \cdot \frac{d}{d y} (\ln (y^{2})) \cdot \ln (y^{2})

Apply the differentiation rule:

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

with f =

y^{2}

ⓘ The derivative of ln(x) is 1/x:

\frac{d}{d y} (\ln (y^{2}))

\frac{d}{d y} (y^{2}) \cdot \frac{1}{y^{2}}

2 \cdot \frac{d}{d y} (y^{2}) \cdot \frac{1}{y^{2}} \cdot \ln (y^{2})

ⓘ Multiplying a fraction by a term multiplies its numerator:

2 \cdot \frac{d}{d y} (y^{2}) \cdot \frac{1}{y^{2}} \cdot \ln (y^{2})

\frac{2 \cdot \frac{d}{d y} (y^{2}) \cdot 1 \cdot \ln (y^{2})}{y^{2}}

\frac{2 \cdot \frac{d}{d y} (y^{2}) \cdot \ln (y^{2})}{y^{2}}

\frac{2 \cdot \frac{d}{d y} (y^{2}) \cdot \ln (y^{2})}{y^{2}}

ⓘ The derivative of x^n is n*x^(n-1):

\frac{d}{d y} (y^{2})

2 y

\frac{2 \cdot 2 y \cdot \ln (y^{2})}{y^{2}}

ⅰ

2 \cdot 2

4

\frac{4 y \cdot \ln (y^{2})}{y^{2}}

We simplify the fraction:

\frac{y}{y^{2}}

by the common factor:

y

to get the new quotient:

\frac{1}{y}

\frac{4 \cdot \ln (y^{2})}{y}

The computed derivative matches the original math expression up to a constant factor. Therefore, the integral is:

ⓘ The derivative of

\frac{1}{12} \cdot {\ln (y^{2})}^{2}

\frac{\ln (y^{2})}{3 y}

, so

\frac{1}{12} \cdot {\ln (y^{2})}^{2}

is an antiderivative of

\frac{\ln (y^{2})}{3 y}

\int \frac{\ln (y^{2})}{3 y} d y

\frac{1}{12} \cdot {\ln (y^{2})}^{2}

\frac{1}{12} \cdot {\ln (y^{2})}^{2}

ⓘ Multiplying a fraction by a term multiplies its numerator:

\frac{1}{12} \cdot {\ln (y^{2})}^{2}

\frac{1 \cdot {\ln (y^{2})}^{2}}{12}

\frac{{\ln (y^{2})}^{2}}{12}

\frac{{\ln (y^{2})}^{2}}{12}

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

\frac{{\ln (y^{2})}^{2}}{12} + C