What is the integral of sqrt(x^2-8x+7) ?

The integral of sqrt(x^2-8x+7) is

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integral of sqrt(x^2-8x+7)

integral

x^{2} - 8 x + 7

9 (- ln (\frac{1}{3} x^{2} - 8 x + 7 + x - 4) + \frac{1}{18} (x - 4) x^{2} - 8 x + 7 + \frac{1}{2} ln (\frac{x ^{2} - 8 x + 7 + x - 4}{3})) + C

Solution steps

\int x^{2} - 8 x + 7 d x

Complete the square

x^{2} - 8 x + 7 : (x - 4)^{2} - 9

= \int (x - 4)^{2} - 9 d x

Apply u-substitution

: \int u^{2} - 9 d u

= \int u^{2} - 9 d u

Apply Trigonometric Substitution

: \int 9 tan^{2} (v) sec (v) d v

= \int 9 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

= 9 \cdot \int tan^{2} (v) sec (v) d v

Rewrite using trig identities

= 9 \cdot \int (- 1 + sec^{2} (v)) sec (v) d v

Expand

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

= 9 \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

= 9 (- \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) ∣

= 9 (- ln ∣ tan (v) + sec (v) ∣ + \frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

Substitute back

= 9 (- ln tan (arcsec (\frac{1}{3} (x - 4))) + sec (arcsec (\frac{1}{3} (x - 4))) + \frac{1}{2} sec (arcsec (\frac{1}{3} (x - 4))) tan (arcsec (\frac{1}{3} (x - 4))) + \frac{1}{2} ln tan (arcsec (\frac{1}{3} (x - 4))) + sec (arcsec (\frac{1}{3} (x - 4))))

Simplify

9 (- ln tan (arcsec (\frac{1}{3} (x - 4))) + sec (arcsec (\frac{1}{3} (x - 4))) + \frac{1}{2} sec (arcsec (\frac{1}{3} (x - 4))) tan (arcsec (\frac{1}{3} (x - 4))) + \frac{1}{2} ln tan (arcsec (\frac{1}{3} (x - 4))) + sec (arcsec (\frac{1}{3} (x - 4)))) : 9 (- ln (\frac{1}{3} x^{2} - 8 x + 7 + x - 4) + \frac{1}{18} (x - 4) x^{2} - 8 x + 7 + \frac{1}{2} ln (\frac{x ^{2} - 8 x + 7 + x - 4}{3}))

= 9 (- ln (\frac{1}{3} x^{2} - 8 x + 7 + x - 4) + \frac{1}{18} (x - 4) x^{2} - 8 x + 7 + \frac{1}{2} ln (\frac{x ^{2} - 8 x + 7 + x - 4}{3}))

Add a constant to the solution

= 9 (- ln (\frac{1}{3} x^{2} - 8 x + 7 + x - 4) + \frac{1}{18} (x - 4) x^{2} - 8 x + 7 + \frac{1}{2} ln (\frac{x ^{2} - 8 x + 7 + x - 4}{3})) + C

\frac{x}{( 1 + x ) ^{3}}

\frac{1}{( x ^{2} + a ^{2} ) ^{\frac{1}{2}}}

\frac{3}{( 2 ( x ) ^{2} - x )}

\frac{sin ^{4} ( x )}{cos ^{3} ( x )}

- 7 x