What is the integral of (ln(t)t^{-1})/(-3) ?

The integral of (ln(t)t^{-1})/(-3) is -1/6 ln^2(t)+C

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integral of (ln(t)t^{-1})/(-3)

\int \frac{ln ( t ) t ^{- 1}}{- 3} d t

- \frac{1}{6} ln^{2} (t) + C

Solution steps

\int \frac{ln ( t ) t ^{- 1}}{- 3} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{- 3} \cdot \int ln (t) t^{- 1} d t

Simplify

= (- \frac{1}{3}) \cdot \int ln (t) t^{- 1} d t

Simplify

ln (t) t^{- 1} : \frac{ln ( t )}{t}

= (- \frac{1}{3}) \cdot \int \frac{ln ( t )}{t} d t

Apply u-substitution

: \int u d u

= (- \frac{1}{3}) \cdot \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= (- \frac{1}{3}) \frac{u ^{2}}{2}

Substitute back

u = ln (t)

= (- \frac{1}{3}) \frac{ln ^{2} ( t )}{2}

Simplify

(- \frac{1}{3}) \frac{ln ^{2} ( t )}{2} : - \frac{1}{6} ln^{2} (t)

= - \frac{1}{6} ln^{2} (t)

Add a constant to the solution

= - \frac{1}{6} ln^{2} (t) + C

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

\int x^{3} 4 - x^{2} d x

\int \frac{5 x ^{3} - 3 x ^{2} + 2}{2 x ^{2}} d x

\int_{0}^{3} 1 + x^{2} d x

x \to 6 + lim (6 x - 6)

What is the integral of (ln(t)t^{-1})/(-3) ?
The integral of (ln(t)t^{-1})/(-3) is -1/6 ln^2(t)+C