What is the integral of (sec^2(8x))/(tan(8x)+5) ?

The integral of (sec^2(8x))/(tan(8x)+5) is 1/8 ln|tan(8x)+5|+C

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integral of (sec^2(8x))/(tan(8x)+5)

\int \frac{sec ^{2} ( 8 x )}{tan ( 8 x ) + 5} d x

\frac{1}{8} ln ∣ tan (8 x) + 5 ∣ + C

Solution steps

\int \frac{sec ^{2} ( 8 x )}{tan ( 8 x ) + 5} 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) + 5

= \frac{1}{8} ln ∣ tan (8 x) + 5 ∣

Add a constant to the solution

= \frac{1}{8} ln ∣ tan (8 x) + 5 ∣ + C

\frac{\partial}{\partial x} (1 (x - y) e^{1 y + 5 x^{2}})

\frac{\partial}{\partial x} (x cos (\frac{x}{y}))

x \to 0 lim (x^{3} - 2 x^{3} - 4 x + 7)

\frac{\partial}{\partial x} (e^{- 3 x^{2} - y^{2}})

\frac{d}{d x} (x^{2} + 4 y^{2} - 4 x y + 2)

What is the integral of (sec^2(8x))/(tan(8x)+5) ?
The integral of (sec^2(8x))/(tan(8x)+5) is 1/8 ln|tan(8x)+5|+C