What is the integral of sqrt(4x^2-12x+7) ?

The integral of sqrt(4x^2-12x+7) is

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integral of sqrt(4x^2-12x+7)

\int 4 x^{2} - 12 x + 7 d x

x 4 x^{2} - 12 x + 7 - \frac{3}{2} 4 x^{2} - 12 x + 7 - \frac{1}{4} (2 x - 3) 4 x^{2} - 12 x + 7 - \frac{1}{2} ln 2 (x^{2} - 3 x + \frac{9}{4}) - 1 + 2 (x - \frac{3}{2}) + C

Solution steps

\int 4 x^{2} - 12 x + 7 d x

Complete the square

4 x^{2} - 12 x + 7 : 4 (x - \frac{3}{2})^{2} - 2

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

Apply u-substitution

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

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

Apply Integration By Parts

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

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

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

= u 4 u^{2} - 2 - (\frac{1}{2} u 2 u^{2} - 1 + \frac{1}{2} ln 2 u^{2} - 1 + 2 u)

Substitute back

u = x - \frac{3}{2}

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

Simplify

(x - \frac{3}{2}) 4 (x - \frac{3}{2})^{2} - 2 - \frac{1}{2} (x - \frac{3}{2}) 2 (x - \frac{3}{2})^{2} - 1 + \frac{1}{2} ln 2 (x - \frac{3}{2})^{2} - 1 + 2 (x - \frac{3}{2}) : x 4 x^{2} - 12 x + 7 - \frac{3}{2} 4 x^{2} - 12 x + 7 - \frac{1}{4} (2 x - 3) 4 x^{2} - 12 x + 7 - \frac{1}{2} ln 2 (x^{2} - 3 x + \frac{9}{4}) - 1 + 2 (x - \frac{3}{2})

= x 4 x^{2} - 12 x + 7 - \frac{3}{2} 4 x^{2} - 12 x + 7 - \frac{1}{4} (2 x - 3) 4 x^{2} - 12 x + 7 - \frac{1}{2} ln 2 (x^{2} - 3 x + \frac{9}{4}) - 1 + 2 (x - \frac{3}{2})

Add a constant to the solution

= x 4 x^{2} - 12 x + 7 - \frac{3}{2} 4 x^{2} - 12 x + 7 - \frac{1}{4} (2 x - 3) 4 x^{2} - 12 x + 7 - \frac{1}{2} ln 2 (x^{2} - 3 x + \frac{9}{4}) - 1 + 2 (x - \frac{3}{2}) + C

y = \frac{x}{9 + x}

x \to 0 lim (\frac{0}{1})

(- 14 x y^{2} + 19 y) + (- 14 x^{2} y + 19 x) y^{^{'}} = 0

\int_{1}^{3} x 4^{(x^{2})} d x

6 t