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

The integral of 1/((x^2-2x-3)^{3/2)} is -(x-1)/(4(x^2-2x-3)^{1/2)}+C

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integral of 1/((x^2-2x-3)^{3/2)}

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

- \frac{x - 1}{4 ( x ^{2} - 2 x - 3 ) ^{\frac{1}{2}}} + C

Solution steps

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

Complete the square

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

= \int \frac{1}{( ( x - 1 ) ^{2} - 4 ) ^{\frac{3}{2}}} d x

Apply u-substitution

: \int \frac{1}{( u ^{2} - 4 ) ^{\frac{3}{2}}} d u

= \int \frac{1}{( u ^{2} - 4 ) ^{\frac{3}{2}}} d u

Apply Trigonometric Substitution

: \int \frac{sec ( v )}{4 tan ^{2} ( v )} d v

= \int \frac{sec ( v )}{4 tan ^{2} ( v )} d v

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{sec ( v )}{tan ^{2} ( v )} d v

Rewrite using trig identities

= \frac{1}{4} \cdot \int \frac{cos ( v )}{sin ^{2} ( v )} d v

Apply u-substitution

: \int \frac{1}{w ^{2}} d w

= \frac{1}{4} \cdot \int \frac{1}{w ^{2}} d w

Apply the Power Rule

: - \frac{1}{w}

= \frac{1}{4} (- \frac{1}{w})

Substitute back

= \frac{1}{4} (- \frac{1}{sin ( arcsec ( \frac{1}{2} ( x - 1 ) ) )})

Simplify

\frac{1}{4} (- \frac{1}{sin ( arcsec ( \frac{1}{2} ( x - 1 ) ) )}) : - \frac{x - 1}{4 ( x ^{2} - 2 x - 3 ) ^{\frac{1}{2}}}

= - \frac{x - 1}{4 ( x ^{2} - 2 x - 3 ) ^{\frac{1}{2}}}

Add a constant to the solution

= - \frac{x - 1}{4 ( x ^{2} - 2 x - 3 ) ^{\frac{1}{2}}} + C

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What is the integral of 1/((x^2-2x-3)^{3/2)} ?
The integral of 1/((x^2-2x-3)^{3/2)} is -(x-1)/(4(x^2-2x-3)^{1/2)}+C