What is the integral of (sec^2(x))/(tan^2(x)-1) ?

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

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integral of (sec^2(x))/(tan^2(x)-1)

\int \frac{sec ^{2} ( x )}{tan ^{2} ( x ) - 1} d x

- \frac{1}{2} ln ∣ tan (x) + 1 ∣ + \frac{1}{2} ln ∣ tan (x) - 1 ∣ + C

Solution steps

\int \frac{sec ^{2} ( x )}{tan ^{2} ( x ) - 1} d x

Apply u-substitution

: \int \frac{1}{u ^{2} - 1} d u

= \int \frac{1}{u ^{2} - 1} d u

Factor

u^{2} - 1 : - (- u^{2} + 1)

= \int \frac{1}{- ( - u ^{2} + 1 )} d u

Take the constant out:

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

= - \int \frac{1}{- u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= - (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Substitute back

u = tan (x)

= - (\frac{ln ∣ tan ( x ) + 1 ∣}{2} - \frac{ln ∣ tan ( x ) - 1 ∣}{2})

Simplify

- (\frac{ln ∣ tan ( x ) + 1 ∣}{2} - \frac{ln ∣ tan ( x ) - 1 ∣}{2}) : - \frac{1}{2} ln ∣ tan (x) + 1 ∣ + \frac{1}{2} ln ∣ tan (x) - 1 ∣

= - \frac{1}{2} ln ∣ tan (x) + 1 ∣ + \frac{1}{2} ln ∣ tan (x) - 1 ∣

Add a constant to the solution

= - \frac{1}{2} ln ∣ tan (x) + 1 ∣ + \frac{1}{2} ln ∣ tan (x) - 1 ∣ + C

x \to 0 lim (csc (x) + cot (x))

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y = (x - 1)^{\frac{3}{2}} + \frac{e ^{x - 2}}{2}

\frac{d}{d t} (t^{2} + 4)

\int \frac{ln ( 10 x )}{3 x} d x

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