What is the integral of 1/(sqrt((x^2-6x+10)^5)) ?

The integral of 1/(sqrt((x^2-6x+10)^5)) is (x-3)/(sqrt(x^2-6x+10))-((x-3)^3)/(3(x^2-6x+10)^{3/2)}+C

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integral of 1/(sqrt((x^2-6x+10)^5))

\int \frac{1}{( x ^{2} - 6 x + 10 ) ^{5}} d x

\frac{x - 3}{x ^{2} - 6 x + 10} - \frac{( x - 3 ) ^{3}}{3 ( x ^{2} - 6 x + 10 ) ^{\frac{3}{2}}} + C

Solution steps

\int \frac{1}{( x ^{2} - 6 x + 10 ) ^{5}} d x

Complete the square

x^{2} - 6 x + 10 : (x - 3)^{2} + 1

= \int \frac{1}{( ( x - 3 ) ^{2} + 1 ) ^{5}} d x

Apply u-substitution

= \int \frac{1}{( u ^{2} + 1 ) ^{\frac{5}{2}}} d u

Apply Trigonometric Substitution

= \int cos^{3} (v) d v

Rewrite using trig identities

= \int (1 - sin^{2} (v)) cos (v) d v

Apply u-substitution

= \int 1 - w^{2} 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

= \int 1 d w - \int w^{2} d w

\int 1 d w = w

\int w^{2} d w = \frac{w ^{3}}{3}

= w - \frac{w ^{3}}{3}

Substitute back

= sin (arctan (x - 3)) - \frac{sin ^{3} ( arctan ( x - 3 ) )}{3}

Simplify

sin (arctan (x - 3)) - \frac{sin ^{3} ( arctan ( x - 3 ) )}{3} : \frac{x - 3}{x ^{2} - 6 x + 10} - \frac{( x - 3 ) ^{3}}{3 ( x ^{2} - 6 x + 10 ) ^{\frac{3}{2}}}

= \frac{x - 3}{x ^{2} - 6 x + 10} - \frac{( x - 3 ) ^{3}}{3 ( x ^{2} - 6 x + 10 ) ^{\frac{3}{2}}}

Add a constant to the solution

= \frac{x - 3}{x ^{2} - 6 x + 10} - \frac{( x - 3 ) ^{3}}{3 ( x ^{2} - 6 x + 10 ) ^{\frac{3}{2}}} + C

What is the integral of 1/(sqrt((x^2-6x+10)^5)) ?
The integral of 1/(sqrt((x^2-6x+10)^5)) is (x-3)/(sqrt(x^2-6x+10))-((x-3)^3)/(3(x^2-6x+10)^{3/2)}+C