What is the integral of 12log_{11}(x) ?

The integral of 12log_{11}(x) is 12(xlog_{11}(x)-1/(ln(11))x)+C

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integral of 12log_{11}(x)

\int 12 lo g_{11} (x) d x

12 (x lo g_{11} (x) - \frac{1}{ln ( 11 )} x) + C

Solution steps

\int 12 lo g_{11} (x) d x

Take the constant out:

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

= 12 \cdot \int lo g_{11} (x) d x

Apply Integration By Parts

: x lo g_{11} (x) - \int \frac{1}{ln ( 11 )} d x

= 12 (x lo g_{11} (x) - \int \frac{1}{ln ( 11 )} d x)

\int \frac{1}{ln ( 11 )} d x = \frac{1}{ln ( 11 )} x

= 12 (x lo g_{11} (x) - \frac{1}{ln ( 11 )} x)

Add a constant to the solution

= 12 (x lo g_{11} (x) - \frac{1}{ln ( 11 )} x) + C

\int \frac{( 3 - x ) ( 5 + x )}{x ^{4}} d x

f (x) = \frac{e ^{3 x}}{x ^{6}}

\frac{d}{d x} (x^{3} + a x^{2} + b x + c)

\frac{d}{d x} (5 x^{2} \cdot ln (x))

(5 x + y e^{\frac{y}{x}}) d x - x e^{\frac{y}{x}} d y = 0, y (1) = 4

What is the integral of 12log_{11}(x) ?
The integral of 12log_{11}(x) is 12(xlog_{11}(x)-1/(ln(11))x)+C