What is the integral from 3 to 4 of 5log_{6}(t) ?

The integral from 3 to 4 of 5log_{6}(t) is (5(8ln(2)-1-3ln(3)))/(ln(6))

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integral from 3 to 4 of 5log_{6}(t)

\int_{3}^{4} 5 lo g_{6} (t) d t

\frac{5 ( 8 ln ( 2 ) - 1 - 3 ln ( 3 ) )}{ln ( 6 )}

Decimal Notation

3.48635 \dots

Solution steps

\int_{3}^{4} 5 lo g_{6} (t) d t

Take the constant out:

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

= 5 \cdot \int_{3}^{4} lo g_{6} (t) d t

Apply Integration By Parts

: [t lo g_{6} (t) - \int \frac{1}{ln ( 6 )} d t]_{3}^{4}

= 5 [t lo g_{6} (t) - \int \frac{1}{ln ( 6 )} d t]_{3}^{4}

\int \frac{1}{ln ( 6 )} d t = \frac{1}{ln ( 6 )} t

= 5 [t lo g_{6} (t) - \frac{1}{ln ( 6 )} t]_{3}^{4}

Join

t lo g_{6} (t) - \frac{1}{ln ( 6 )} t : \frac{- t + t ln ( t )}{ln ( 6 )}

= 5 [\frac{- t + t ln ( t )}{ln ( 6 )}]_{3}^{4}

Compute the boundaries:

\frac{8 ln ( 2 ) - 1 - 3 ln ( 3 )}{ln ( 6 )}

= 5 \cdot \frac{8 ln ( 2 ) - 1 - 3 ln ( 3 )}{ln ( 6 )}

Simplify

= \frac{5 ( 8 ln ( 2 ) - 1 - 3 ln ( 3 ) )}{ln ( 6 )}

\int \frac{1}{3 - y} d y

x \to 4 lim (\frac{x ^{2} - 2}{x - 4})

\frac{d y}{d x} - y = 1

\int 3 cos (x) + e^{- x} + 4 d x

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

What is the integral from 3 to 4 of 5log_{6}(t) ?
The integral from 3 to 4 of 5log_{6}(t) is (5(8ln(2)-1-3ln(3)))/(ln(6))