What is the integral of 1/((x-1)sqrt(x^2-2x-3)) ?

The integral of 1/((x-1)sqrt(x^2-2x-3)) is 1/2 arcsec(1/2 (x-1))+C

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integral of 1/((x-1)sqrt(x^2-2x-3))

\int \frac{1}{( x - 1 ) x ^{2} - 2 x - 3} d x

\frac{1}{2} arcsec (\frac{1}{2} (x - 1)) + C

Solution steps

\int \frac{1}{( x - 1 ) x ^{2} - 2 x - 3} d x

Complete the square

x^{2} - 2 x - 3 : (x - 1)^{2} - 4

= \int \frac{1}{( x - 1 ) ( x - 1 ) ^{2} - 4} d x

Apply u-substitution

: \int \frac{1}{u u ^{2} - 4} d u

= \int \frac{1}{u u ^{2} - 4} d u

Apply Trigonometric Substitution

: \int \frac{1}{2} d v

= \int \frac{1}{2} d v

Take the constant out:

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

= \frac{1}{2} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot 1 \cdot v

Substitute back

= \frac{1}{2} \cdot 1 \cdot arcsec (\frac{1}{2} (x - 1))

Simplify

= \frac{1}{2} arcsec (\frac{1}{2} (x - 1))

Add a constant to the solution

= \frac{1}{2} arcsec (\frac{1}{2} (x - 1)) + C

\frac{d}{d x} (\frac{4 x}{3 x + 1})

x \to 5 lim ((3 + \frac{1}{x})^{3})

y^{^{'}} = \frac{4 x}{e ^{x}}

x^{2} (1 - 9 x)

f (x) = x^{4} - 20 x^{2} + 64

What is the integral of 1/((x-1)sqrt(x^2-2x-3)) ?
The integral of 1/((x-1)sqrt(x^2-2x-3)) is 1/2 arcsec(1/2 (x-1))+C