What is the integral of 1/((4t-1)ln(8t-2)) ?

The integral of 1/((4t-1)ln(8t-2)) is 1/4 ln|ln(2(4t-1))|+C

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integral of 1/((4t-1)ln(8t-2))

\int \frac{1}{( 4 t - 1 ) ln ( 8 t - 2 )} d t

\frac{1}{4} ln ∣ ln (2 (4 t - 1)) ∣ + C

Solution steps

\int \frac{1}{( 4 t - 1 ) ln ( 8 t - 2 )} d t

Apply u-substitution

: \int \frac{1}{4 u ln ( 2 u )} d u

= \int \frac{1}{4 u ln ( 2 u )} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{u ln ( 2 u )} d u

Apply u-substitution

: \int \frac{1}{v ln ( v )} d v

= \frac{1}{4} \cdot \int \frac{1}{v ln ( v )} d v

Apply u-substitution

: \int \frac{1}{w} d w

= \frac{1}{4} \cdot \int \frac{1}{w} d w

Use the common integral:

\int \frac{1}{w} d w = ln (∣ w ∣)

= \frac{1}{4} ln ∣ w ∣

Substitute back

= \frac{1}{4} ln ∣ ln (2 (4 t - 1)) ∣

Add a constant to the solution

= \frac{1}{4} ln ∣ ln (2 (4 t - 1)) ∣ + C

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What is the integral of 1/((4t-1)ln(8t-2)) ?
The integral of 1/((4t-1)ln(8t-2)) is 1/4 ln|ln(2(4t-1))|+C