What is the integral of (sec^2(5x))/(tan(5x)-4) ?

The integral of (sec^2(5x))/(tan(5x)-4) is 1/5 ln|tan(5x)-4|+C

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integral of (sec^2(5x))/(tan(5x)-4)

\int \frac{sec ^{2} ( 5 x )}{tan ( 5 x ) - 4} d x

\frac{1}{5} ln ∣ tan (5 x) - 4 ∣ + C

Solution steps

\int \frac{sec ^{2} ( 5 x )}{tan ( 5 x ) - 4} d x

Apply u-substitution

: \int \frac{1}{5 u} d u

= \int \frac{1}{5 u} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{5} ln ∣ u ∣

Substitute back

u = tan (5 x) - 4

= \frac{1}{5} ln ∣ tan (5 x) - 4 ∣

Add a constant to the solution

= \frac{1}{5} ln ∣ tan (5 x) - 4 ∣ + C

y^{^{'}} = 4 y + a

f (x) = 2 x - x

\int_{0}^{2 - 2 x} 1 + 3 x + y d y

\int (5 x^{2} + 1) (5 x^{3} + 3 x - 8)^{6} d x

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

What is the integral of (sec^2(5x))/(tan(5x)-4) ?
The integral of (sec^2(5x))/(tan(5x)-4) is 1/5 ln|tan(5x)-4|+C