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

Solution

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

\frac{1}{8 2} (2 arcsec (\frac{1}{2} (x - 1)) + sin (2 arcsec (\frac{1}{2} (x - 1)))) + C

Solution steps

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

Apply u-substitution

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

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

Apply Integral Substitution

= \int \frac{8}{v ^{3} v ^{2} - 8} d v

Take the constant out:

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

= 8 \cdot \int \frac{1}{v ^{3} v ^{2} - 8} d v

Apply Trigonometric Substitution

: \int \frac{1}{16 2 sec ^{2} ( w )} d w

= 8 \cdot \int \frac{1}{16 2 sec ^{2} ( w )} d w

Take the constant out:

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

= 8 \cdot \frac{1}{16 2} \cdot \int \frac{1}{sec ^{2} ( w )} d w

Rewrite using trig identities

= 8 \cdot \frac{1}{16 2} \cdot \int \frac{1 + cos ( 2 w )}{2} d w

Take the constant out:

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

= 8 \cdot \frac{1}{16 2} \cdot \frac{1}{2} \cdot \int 1 + cos (2 w) d w

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 8 \cdot \frac{1}{16 2} \cdot \frac{1}{2} (\int 1 d w + \int cos (2 w) d w)

\int 1 d w = w

\int cos (2 w) d w = \frac{1}{2} sin (2 w)

= 8 \cdot \frac{1}{16 2} \cdot \frac{1}{2} (w + \frac{1}{2} sin (2 w))

Substitute back

= 8 \cdot \frac{1}{16 2} \cdot \frac{1}{2} (arcsec (\frac{1}{2 2} \cdot 2 (x - 1)) + \frac{1}{2} sin (2 arcsec (\frac{1}{2 2} \cdot 2 (x - 1))))

Simplify

8 \cdot \frac{1}{16 2} \cdot \frac{1}{2} (arcsec (\frac{1}{2 2} \cdot 2 (x - 1)) + \frac{1}{2} sin (2 arcsec (\frac{1}{2 2} \cdot 2 (x - 1)))) : \frac{1}{8 2} (2 arcsec (\frac{1}{2} (x - 1)) + sin (2 arcsec (\frac{1}{2} (x - 1))))

= \frac{1}{8 2} (2 arcsec (\frac{1}{2} (x - 1)) + sin (2 arcsec (\frac{1}{2} (x - 1))))

Add a constant to the solution

= \frac{1}{8 2} (2 arcsec (\frac{1}{2} (x - 1)) + sin (2 arcsec (\frac{1}{2} (x - 1)))) + C

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