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

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

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

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

\frac{x - 3}{( x ^{2} - 6 x + 10 ) ^{\frac{1}{2}}} - \frac{( x - 3 ) ^{3}}{3 ( x ^{2} - 6 x + 10 ) ^{\frac{3}{2}}} + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

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

= \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

= \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{1}{2}}} - \frac{( x - 3 ) ^{3}}{3 ( x ^{2} - 6 x + 10 ) ^{\frac{3}{2}}}

= \frac{x - 3}{( x ^{2} - 6 x + 10 ) ^{\frac{1}{2}}} - \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{1}{2}}} - \frac{( x - 3 ) ^{3}}{3 ( x ^{2} - 6 x + 10 ) ^{\frac{3}{2}}} + C

\int \frac{cos ( 2 x )}{sin ^{2} ( x )} d x

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

\int \frac{1}{3 + 2 cos ( x )} d x

\int sin (2 x) cos (2 x) d x

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