What is the integral from 1 to 8 of 2log_{4}(t+3) ?

The integral from 1 to 8 of 2log_{4}(t+3) is 16log_{4}(11)+(3ln(11)+ln(27/8)-3ln(6)-7)/(ln(2))-2

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integral from 1 to 8 of 2log_{4}(t+3)

\int_{1}^{8} 2 lo g_{4} (t + 3) d t

16 lo g_{4} (11) + \frac{3 ln ( 11 ) + ln ( \frac{27}{8} ) - 3 ln ( 6 ) - 7}{ln ( 2 )} - 2

Decimal

19.95488 \dots

Solution steps

\int_{1}^{8} 2 lo g_{4} (t + 3) d t

Take the constant out:

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

= 2 \cdot \int_{1}^{8} lo g_{4} (t + 3) d t

Multiply by the conjugate of

t + 3 : \frac{t ^{2} - 9}{t - 3}

= 2 \cdot \int_{1}^{8} lo g_{4} (\frac{t ^{2} - 9}{t - 3}) d t

Find undefined (singularity) points:

t = 3

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= 2 (\int_{1}^{3} lo g_{4} (\frac{t ^{2} - 9}{t - 3}) d t + \int_{3}^{8} lo g_{4} (\frac{t ^{2} - 9}{t - 3}) d t)

\int_{1}^{3} lo g_{4} (\frac{t ^{2} - 9}{t - 3}) d t = 3 lo g_{4} (6) + \frac{ln ( \frac{27}{8} ) - 2}{2 ln ( 2 )} - 1

\int_{3}^{8} lo g_{4} (\frac{t ^{2} - 9}{t - 3}) d t = - 3 lo g_{4} (6) + 8 lo g_{4} (11) + \frac{3 ln ( 11 ) - 3 ln ( 6 ) - 5}{2 ln ( 2 )}

= 2 (3 lo g_{4} (6) + \frac{ln ( \frac{27}{8} ) - 2}{2 ln ( 2 )} - 1 - 3 lo g_{4} (6) + 8 lo g_{4} (11) + \frac{3 ln ( 11 ) - 3 ln ( 6 ) - 5}{2 ln ( 2 )})

Simplify

2 (3 lo g_{4} (6) + \frac{ln ( \frac{27}{8} ) - 2}{2 ln ( 2 )} - 1 - 3 lo g_{4} (6) + 8 lo g_{4} (11) + \frac{3 ln ( 11 ) - 3 ln ( 6 ) - 5}{2 ln ( 2 )}) : 16 lo g_{4} (11) + \frac{3 ln ( 11 ) + ln ( \frac{27}{8} ) - 3 ln ( 6 ) - 7}{ln ( 2 )} - 2

= 16 lo g_{4} (11) + \frac{3 ln ( 11 ) + ln ( \frac{27}{8} ) - 3 ln ( 6 ) - 7}{ln ( 2 )} - 2

What is the integral from 1 to 8 of 2log_{4}(t+3) ?
The integral from 1 to 8 of 2log_{4}(t+3) is 16log_{4}(11)+(3ln(11)+ln(27/8)-3ln(6)-7)/(ln(2))-2