What is the integral of 11log_{14}(t) ?

The integral of 11log_{14}(t) is 11(tlog_{14}(t)-1/(ln(14))t)+C

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integral of 11log_{14}(t)

\int 11 lo g_{14} (t) d t

11 (t lo g_{14} (t) - \frac{1}{ln ( 14 )} t) + C

Solution steps

\int 11 lo g_{14} (t) d t

Take the constant out:

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

= 11 \cdot \int lo g_{14} (t) d t

Apply Integration By Parts

: t lo g_{14} (t) - \int \frac{1}{ln ( 14 )} d t

= 11 (t lo g_{14} (t) - \int \frac{1}{ln ( 14 )} d t)

\int \frac{1}{ln ( 14 )} d t = \frac{1}{ln ( 14 )} t

= 11 (t lo g_{14} (t) - \frac{1}{ln ( 14 )} t)

Add a constant to the solution

= 11 (t lo g_{14} (t) - \frac{1}{ln ( 14 )} t) + C

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What is the integral of 11log_{14}(t) ?
The integral of 11log_{14}(t) is 11(tlog_{14}(t)-1/(ln(14))t)+C