What is the integral of 1/(sqrt((x^2-6x+25)^3)) ?

The integral of 1/(sqrt((x^2-6x+25)^3)) is (x-3)/(16sqrt(x^2-6x+25))+C

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integral of 1/(sqrt((x^2-6x+25)^3))

\int \frac{1}{( x ^{2} - 6 x + 25 ) ^{3}} d x

\frac{x - 3}{16 x ^{2} - 6 x + 25} + C

Solution steps

\int \frac{1}{( x ^{2} - 6 x + 25 ) ^{3}} d x

(x^{2} - 6 x + 25)^{3} = (x^{2} - 6 x + 25)^{\frac{3}{2}},

assuming

(x^{2} - 6 x + 25) \geq 0

= \int \frac{1}{( x ^{2} - 6 x + 25 ) ^{\frac{3}{2}}} d x

Complete the square

x^{2} - 6 x + 25 : (x - 3)^{2} + 16

= \int \frac{1}{( ( x - 3 ) ^{2} + 16 ) ^{\frac{3}{2}}} d x

Apply u-substitution

: \int \frac{1}{( u ^{2} + 16 ) ^{\frac{3}{2}}} d u

= \int \frac{1}{( u ^{2} + 16 ) ^{\frac{3}{2}}} d u

Apply Trigonometric Substitution

: \int \frac{1}{16 sec ( v )} d v

= \int \frac{1}{16 sec ( v )} d v

Take the constant out:

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

= \frac{1}{16} \cdot \int \frac{1}{sec ( v )} d v

Rewrite using trig identities

= \frac{1}{16} \cdot \int cos (v) d v

Use the common integral:

\int cos (v) d v = sin (v)

= \frac{1}{16} sin (v)

Substitute back

= \frac{1}{16} sin (arctan (\frac{1}{4} (x - 3)))

Simplify

\frac{1}{16} sin (arctan (\frac{1}{4} (x - 3))) : \frac{x - 3}{16 x ^{2} - 6 x + 25}

= \frac{x - 3}{16 x ^{2} - 6 x + 25}

Add a constant to the solution

= \frac{x - 3}{16 x ^{2} - 6 x + 25} + C

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What is the integral of 1/(sqrt((x^2-6x+25)^3)) ?
The integral of 1/(sqrt((x^2-6x+25)^3)) is (x-3)/(16sqrt(x^2-6x+25))+C