What is the integral of (x^2-6x+9)/(sqrt(x^2-6x+58)) ?

The integral of (x^2-6x+9)/(sqrt(x^2-6x+58)) is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^2-6x+9)/(sqrt(x^2-6x+58))

\int \frac{x ^{2} - 6 x + 9}{x ^{2} - 6 x + 58} d x

49 (- ln (\frac{1}{7} x - 3 + x^{2} - 6 x + 58) + \frac{1}{98} (x - 3) x^{2} - 6 x + 58 + \frac{1}{2} ln (\frac{x - 3 + x ^{2} - 6 x + 58}{7})) + C

Solution steps

\int \frac{x ^{2} - 6 x + 9}{x ^{2} - 6 x + 58} d x

Complete the square

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

= \int \frac{( x - 3 ) ^{2}}{x ^{2} - 6 x + 58} d x

Apply u-substitution

: \int \frac{u ^{2}}{u ^{2} + 49} d u

= \int \frac{u ^{2}}{u ^{2} + 49} d u

Apply Trigonometric Substitution

: \int 49 tan^{2} (v) sec (v) d v

= \int 49 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

= 49 \cdot \int tan^{2} (v) sec (v) d v

Rewrite using trig identities

= 49 \cdot \int (- 1 + sec^{2} (v)) sec (v) d v

Expand

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

= 49 \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

= 49 (- \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) ∣

= 49 (- ln ∣ tan (v) + sec (v) ∣ + \frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

Substitute back

= 49 (- ln tan (arctan (\frac{1}{7} (x - 3))) + sec (arctan (\frac{1}{7} (x - 3))) + \frac{1}{2} sec (arctan (\frac{1}{7} (x - 3))) tan (arctan (\frac{1}{7} (x - 3))) + \frac{1}{2} ln tan (arctan (\frac{1}{7} (x - 3))) + sec (arctan (\frac{1}{7} (x - 3))))

Simplify

49 (- ln tan (arctan (\frac{1}{7} (x - 3))) + sec (arctan (\frac{1}{7} (x - 3))) + \frac{1}{2} sec (arctan (\frac{1}{7} (x - 3))) tan (arctan (\frac{1}{7} (x - 3))) + \frac{1}{2} ln tan (arctan (\frac{1}{7} (x - 3))) + sec (arctan (\frac{1}{7} (x - 3)))) : 49 (- ln (\frac{1}{7} x - 3 + x^{2} - 6 x + 58) + \frac{1}{98} (x - 3) x^{2} - 6 x + 58 + \frac{1}{2} ln (\frac{x - 3 + x ^{2} - 6 x + 58}{7}))

= 49 (- ln (\frac{1}{7} x - 3 + x^{2} - 6 x + 58) + \frac{1}{98} (x - 3) x^{2} - 6 x + 58 + \frac{1}{2} ln (\frac{x - 3 + x ^{2} - 6 x + 58}{7}))

Add a constant to the solution

= 49 (- ln (\frac{1}{7} x - 3 + x^{2} - 6 x + 58) + \frac{1}{98} (x - 3) x^{2} - 6 x + 58 + \frac{1}{2} ln (\frac{x - 3 + x ^{2} - 6 x + 58}{7})) + C

\int (2 x^{2} - \frac{3}{4} x) d x

\int \frac{x}{( x ^{2} - 2 ) x ^{4} - 4 x ^{2} + 5} d x

\int (\frac{x ^{3}}{3} - \frac{x ^{4}}{2} + \frac{x ^{5}}{5}) d x

\int \frac{1}{x + 3 x - 4} d x

\int \frac{1}{2} cot^{2} (x) csc^{4} (x) d x

What is the integral of (x^2-6x+9)/(sqrt(x^2-6x+58)) ?
The integral of (x^2-6x+9)/(sqrt(x^2-6x+58)) is