What is the integral from 1 to x of sqrt(t^2-t) ?

The integral from 1 to x of sqrt(t^2-t) is (-ln|2x+2sqrt(x(x-1))-1|+2(2x-1)sqrt(x(x-1)))/8

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integral from 1 to x of sqrt(t^2-t)

\int_{1}^{x} t^{2} - t d t

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

Solution steps

\int_{1}^{x} t^{2} - t d t

Complete the square

t^{2} - t : (t - \frac{1}{2})^{2} - \frac{1}{4}

= \int_{1}^{x} (t - \frac{1}{2})^{2} - \frac{1}{4} d t

Apply u-substitution

: \int_{\frac{1}{2}}^{x - \frac{1}{2}} \frac{4 u ^{2} - 1}{2} d u

= \int_{\frac{1}{2}}^{x - \frac{1}{2}} \frac{4 u ^{2} - 1}{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_{\frac{1}{2}}^{x - \frac{1}{2}} 4 u^{2} - 1 d u

Apply Trigonometric Substitution

: \int_{0}^{arcsec (2 x - 1)} \frac{1}{2} tan^{2} (v) sec (v) d v

= \frac{1}{2} \cdot \int_{0}^{arcsec (2 x - 1)} \frac{1}{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{1}{2} \cdot \int_{0}^{arcsec (2 x - 1)} tan^{2} (v) sec (v) d v

Rewrite using trig identities

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{0}^{arcsec (2 x - 1)} (- 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{1}{2} \cdot \int_{0}^{arcsec (2 x - 1)} - 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{1}{2} (- \int_{0}^{arcsec (2 x - 1)} sec (v) d v + \int_{0}^{arcsec (2 x - 1)} sec^{3} (v) d v)

\int_{0}^{arcsec (2 x - 1)} sec (v) d v = ln 2 x + 2 x (x - 1) - 1

\int_{0}^{arcsec (2 x - 1)} sec^{3} (v) d v = (2 x - 1) x (x - 1) + \frac{1}{2} ln 2 x + 2 x (x - 1) - 1

= \frac{1}{2} \cdot \frac{1}{2} (- ln 2 x + 2 x (x - 1) - 1 + (2 x - 1) x (x - 1) + \frac{1}{2} ln 2 x + 2 x (x - 1) - 1)

Simplify

\frac{1}{2} \cdot \frac{1}{2} (- ln 2 x + 2 x (x - 1) - 1 + (2 x - 1) x (x - 1) + \frac{1}{2} ln 2 x + 2 x (x - 1) - 1) : \frac{- ln 2 x + 2 x ( x - 1 ) - 1 + 2 ( 2 x - 1 ) x ( x - 1 )}{8}

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

\int_{1}^{e} \frac{10 ln ( x )}{x} d x

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\int_{a}^{10 b} 2 s d s

\int_{- 1}^{1} 3 x^{5} - 3 x^{2} - 3 x + 3 d x

What is the integral from 1 to x of sqrt(t^2-t) ?
The integral from 1 to x of sqrt(t^2-t) is (-ln|2x+2sqrt(x(x-1))-1|+2(2x-1)sqrt(x(x-1)))/8