What is the integral of sqrt(x^2-5x) ?

The integral of sqrt(x^2-5x) is 1/8 (2(2x-5)sqrt(x(x-5))-25ln((|2x+2sqrt(x(x-5))-5|)/5))+C

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integral of sqrt(x^2-5x)

\int x^{2} - 5 x d x

\frac{1}{8} 2 (2 x - 5) x (x - 5) - 25 ln \frac{2 x + 2 x ( x - 5 ) - 5}{5} + C

Solution steps

\int x^{2} - 5 x d x

Complete the square

x^{2} - 5 x : (x - \frac{5}{2})^{2} - \frac{25}{4}

= \int (x - \frac{5}{2})^{2} - \frac{25}{4} d x

Apply u-substitution

: \int \frac{4 u ^{2} - 25}{2} d u

= \int \frac{4 u ^{2} - 25}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int 4 u^{2} - 25 d u

Apply Trigonometric Substitution

: \int \frac{25}{2} tan^{2} (v) sec (v) d v

= \frac{1}{2} \cdot \int \frac{25}{2} 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

= \frac{1}{2} \cdot \frac{25}{2} \cdot \int tan^{2} (v) sec (v) d v

Rewrite using trig identities

= \frac{1}{2} \cdot \frac{25}{2} \cdot \int (- 1 + sec^{2} (v)) sec (v) d v

Expand

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

= \frac{1}{2} \cdot \frac{25}{2} \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

= \frac{1}{2} \cdot \frac{25}{2} (- \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) ∣

= \frac{1}{2} \cdot \frac{25}{2} (- ln ∣ tan (v) + sec (v) ∣ + \frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

Substitute back

= \frac{1}{2} \cdot \frac{25}{2} (- ln tan (arcsec (\frac{2}{5} (x - \frac{5}{2}))) + sec (arcsec (\frac{2}{5} (x - \frac{5}{2}))) + \frac{1}{2} sec (arcsec (\frac{2}{5} (x - \frac{5}{2}))) tan (arcsec (\frac{2}{5} (x - \frac{5}{2}))) + \frac{1}{2} ln tan (arcsec (\frac{2}{5} (x - \frac{5}{2}))) + sec (arcsec (\frac{2}{5} (x - \frac{5}{2}))))

Simplify

\frac{1}{2} \cdot \frac{25}{2} (- ln tan (arcsec (\frac{2}{5} (x - \frac{5}{2}))) + sec (arcsec (\frac{2}{5} (x - \frac{5}{2}))) + \frac{1}{2} sec (arcsec (\frac{2}{5} (x - \frac{5}{2}))) tan (arcsec (\frac{2}{5} (x - \frac{5}{2}))) + \frac{1}{2} ln tan (arcsec (\frac{2}{5} (x - \frac{5}{2}))) + sec (arcsec (\frac{2}{5} (x - \frac{5}{2})))) : \frac{1}{8} 2 (2 x - 5) x (x - 5) - 25 ln \frac{2 x + 2 x ( x - 5 ) - 5}{5}

= \frac{1}{8} 2 (2 x - 5) x (x - 5) - 25 ln \frac{2 x + 2 x ( x - 5 ) - 5}{5}

Add a constant to the solution

= \frac{1}{8} 2 (2 x - 5) x (x - 5) - 25 ln \frac{2 x + 2 x ( x - 5 ) - 5}{5} + C

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What is the integral of sqrt(x^2-5x) ?
The integral of sqrt(x^2-5x) is 1/8 (2(2x-5)sqrt(x(x-5))-25ln((|2x+2sqrt(x(x-5))-5|)/5))+C