What is the integral of sqrt(x^2-3x+2) ?

The integral of sqrt(x^2-3x+2) is 1/8 (4xsqrt(x^2-3x+2)-6sqrt(x^2-3x+2)-ln|2x-3+2sqrt(x^2-3x+2)|)+C

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integral of sqrt(x^2-3x+2)

\int x^{2} - 3 x + 2 d x

\frac{1}{8} (4 x x^{2} - 3 x + 2 - 6 x^{2} - 3 x + 2 - ln 2 x - 3 + 2 x^{2} - 3 x + 2) + C

Solution steps

\int x^{2} - 3 x + 2 d x

Complete the square

x^{2} - 3 x + 2 : (x - \frac{3}{2})^{2} - \frac{1}{4}

= \int (x - \frac{3}{2})^{2} - \frac{1}{4} d x

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int \frac{1}{2} tan^{2} (v) sec (v) d v

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

Rewrite using trig identities

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

Simplify

\frac{1}{2} \cdot \frac{1}{2} (- ln tan (arcsec (2 (x - \frac{3}{2}))) + sec (arcsec (2 (x - \frac{3}{2}))) + \frac{1}{2} sec (arcsec (2 (x - \frac{3}{2}))) tan (arcsec (2 (x - \frac{3}{2}))) + \frac{1}{2} ln tan (arcsec (2 (x - \frac{3}{2}))) + sec (arcsec (2 (x - \frac{3}{2})))) : \frac{1}{8} (4 x x^{2} - 3 x + 2 - 6 x^{2} - 3 x + 2 - ln 2 x - 3 + 2 x^{2} - 3 x + 2)

= \frac{1}{8} (4 x x^{2} - 3 x + 2 - 6 x^{2} - 3 x + 2 - ln 2 x - 3 + 2 x^{2} - 3 x + 2)

Add a constant to the solution

= \frac{1}{8} (4 x x^{2} - 3 x + 2 - 6 x^{2} - 3 x + 2 - ln 2 x - 3 + 2 x^{2} - 3 x + 2) + C

\frac{\partial}{\partial x} (\frac{x}{x + y ^{2}})

y^{^{'''}} - 2 y^{^{''}} - y^{^{'}} + 2 y = 0

3 r = \frac{d r}{d θ} - θ^{3}

\int cos (\frac{1}{5} x) d x

\int sin (x) + 1 d x

What is the integral of sqrt(x^2-3x+2) ?
The integral of sqrt(x^2-3x+2) is 1/8 (4xsqrt(x^2-3x+2)-6sqrt(x^2-3x+2)-ln|2x-3+2sqrt(x^2-3x+2)|)+C