What is the integral of (sec^2(8x))/(tan(8x)-6) ?

The integral of (sec^2(8x))/(tan(8x)-6) is 1/8 ln|tan(8x)-6|+C

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integral of (sec^2(8x))/(tan(8x)-6)

\int \frac{sec ^{2} ( 8 x )}{tan ( 8 x ) - 6} d x

\frac{1}{8} ln ∣ tan (8 x) - 6 ∣ + C

Solution steps

\int \frac{sec ^{2} ( 8 x )}{tan ( 8 x ) - 6} d x

Apply u-substitution

: \int \frac{1}{8 u} d u

= \int \frac{1}{8 u} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{8} ln ∣ u ∣

Substitute back

u = tan (8 x) - 6

= \frac{1}{8} ln ∣ tan (8 x) - 6 ∣

Add a constant to the solution

= \frac{1}{8} ln ∣ tan (8 x) - 6 ∣ + C

\frac{d y}{d x} = x e^{x^{2} - 4}

n = 1 \sum \infty (- 8)^{- n}

\frac{\partial}{\partial z} (e^{x y} sin (z))

\int (x^{2} + 3 x)^{4} (2 x + 3) d x

4 - 2 e^{- 4 t}

What is the integral of (sec^2(8x))/(tan(8x)-6) ?
The integral of (sec^2(8x))/(tan(8x)-6) is 1/8 ln|tan(8x)-6|+C