What is the integral from 4 to 7 of-9log_{3}(t) ?

The integral from 4 to 7 of-9log_{3}(t) is -(9(7ln(7)-3-8ln(2)))/(ln(3))

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integral from 4 to 7 of-9log_{3}(t)

\int_{4}^{7} - 9 lo g_{3} (t) d t

- \frac{9 ( 7 ln ( 7 ) - 3 - 8 ln ( 2 ) )}{ln ( 3 )}

Solution steps

\int_{4}^{7} - 9 lo g_{3} (t) d t

Take the constant out:

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

= - 9 \cdot \int_{4}^{7} lo g_{3} (t) d t

Apply Integration By Parts

: [t lo g_{3} (t) - \int \frac{1}{ln ( 3 )} d t]_{4}^{7}

= - 9 [t lo g_{3} (t) - \int \frac{1}{ln ( 3 )} d t]_{4}^{7}

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

= - 9 [t lo g_{3} (t) - \frac{1}{ln ( 3 )} t]_{4}^{7}

Join

t lo g_{3} (t) - \frac{1}{ln ( 3 )} t : \frac{- t + t ln ( t )}{ln ( 3 )}

= - 9 [\frac{- t + t ln ( t )}{ln ( 3 )}]_{4}^{7}

Compute the boundaries:

\frac{7 ln ( 7 ) - 3 - 8 ln ( 2 )}{ln ( 3 )}

= - 9 \cdot \frac{7 ln ( 7 ) - 3 - 8 ln ( 2 )}{ln ( 3 )}

Simplify

= - \frac{9 ( 7 ln ( 7 ) - 3 - 8 ln ( 2 ) )}{ln ( 3 )}

\int_{1}^{4} (3 x^{2} + \frac{4}{x ^{2}}) d x

\int_{1}^{4} (\frac{1}{x}) d x

\int_{0}^{x} ln (1 + t^{2}) d t

\int_{3}^{7} 10 x d x

\int_{1}^{2} \frac{1}{t ^{2}} d t

What is the integral from 4 to 7 of-9log_{3}(t) ?
The integral from 4 to 7 of-9log_{3}(t) is -(9(7ln(7)-3-8ln(2)))/(ln(3))