What is the integral of (x^2-6x+9)/(sqrt(x^2-6x+25)) ?

The integral of (x^2-6x+9)/(sqrt(x^2-6x+25)) is

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integral of (x^2-6x+9)/(sqrt(x^2-6x+25))

\int \frac{x ^{2} - 6 x + 9}{x ^{2} - 6 x + 25} d x

16 (- ln (\frac{1}{4} x - 3 + x^{2} - 6 x + 25) + \frac{1}{32} (x - 3) x^{2} - 6 x + 25 + \frac{1}{2} ln (\frac{x - 3 + x ^{2} - 6 x + 25}{4})) + C

Solution steps

\int \frac{x ^{2} - 6 x + 9}{x ^{2} - 6 x + 25} d x

Complete the square

x^{2} - 6 x + 9 : (x - 3)^{2}

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

Apply u-substitution

: \int \frac{u ^{2}}{u ^{2} + 16} d u

= \int \frac{u ^{2}}{u ^{2} + 16} d u

Apply Trigonometric Substitution

: \int 16 tan^{2} (v) sec (v) d v

= \int 16 tan^{2} (v) sec (v) d v

Take the constant out:

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

= 16 \cdot \int tan^{2} (v) sec (v) d v

Rewrite using trig identities

= 16 \cdot \int (- 1 + sec^{2} (v)) sec (v) d v

Expand

(- 1 + sec^{2} (v)) sec (v) : - sec (v) + sec^{3} (v)

= 16 \cdot \int - sec (v) + sec^{3} (v) d v

Apply the Sum Rule:

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

= 16 (- \int sec (v) d v + \int sec^{3} (v) d v)

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

\int sec^{3} (v) d v = \frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

= 16 (- ln ∣ tan (v) + sec (v) ∣ + \frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

Substitute back

= 16 (- ln tan (arctan (\frac{1}{4} (x - 3))) + sec (arctan (\frac{1}{4} (x - 3))) + \frac{1}{2} sec (arctan (\frac{1}{4} (x - 3))) tan (arctan (\frac{1}{4} (x - 3))) + \frac{1}{2} ln tan (arctan (\frac{1}{4} (x - 3))) + sec (arctan (\frac{1}{4} (x - 3))))

Simplify

16 (- ln tan (arctan (\frac{1}{4} (x - 3))) + sec (arctan (\frac{1}{4} (x - 3))) + \frac{1}{2} sec (arctan (\frac{1}{4} (x - 3))) tan (arctan (\frac{1}{4} (x - 3))) + \frac{1}{2} ln tan (arctan (\frac{1}{4} (x - 3))) + sec (arctan (\frac{1}{4} (x - 3)))) : 16 (- ln (\frac{1}{4} x - 3 + x^{2} - 6 x + 25) + \frac{1}{32} (x - 3) x^{2} - 6 x + 25 + \frac{1}{2} ln (\frac{x - 3 + x ^{2} - 6 x + 25}{4}))

= 16 (- ln (\frac{1}{4} x - 3 + x^{2} - 6 x + 25) + \frac{1}{32} (x - 3) x^{2} - 6 x + 25 + \frac{1}{2} ln (\frac{x - 3 + x ^{2} - 6 x + 25}{4}))

Add a constant to the solution

= 16 (- ln (\frac{1}{4} x - 3 + x^{2} - 6 x + 25) + \frac{1}{32} (x - 3) x^{2} - 6 x + 25 + \frac{1}{2} ln (\frac{x - 3 + x ^{2} - 6 x + 25}{4})) + C

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What is the integral of (x^2-6x+9)/(sqrt(x^2-6x+25)) ?
The integral of (x^2-6x+9)/(sqrt(x^2-6x+25)) is