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

The integral of xsqrt(x^2-2x+2) is

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integral of xsqrt(x^2-2x+2)

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

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

Solution steps

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

Complete the square

x^{2} - 2 x + 2 : (x - 1)^{2} + 1

= \int x (x - 1)^{2} + 1 d x

Apply u-substitution

= \int (u + 1) u^{2} + 1 d u

Apply Integration By Parts

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

\int \frac{1}{2} (u 1 + u^{2} + ln u + 1 + u^{2}) d u = \frac{1}{2} (\frac{1}{3} (1 + u^{2})^{\frac{3}{2}} + u ln u + 1 + u^{2} - 1 + u^{2})

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

Substitute back

u = x - 1

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

Simplify

(x - 1 + 1) (\frac{1}{2} (x - 1) 1 + (x - 1)^{2} + \frac{1}{2} ln x - 1 + 1 + (x - 1)^{2}) - \frac{1}{2} (\frac{1}{3} (1 + (x - 1)^{2})^{\frac{3}{2}} + (x - 1) ln x - 1 + 1 + (x - 1)^{2} - 1 + (x - 1)^{2}) : x (\frac{1}{2} (x - 1) x^{2} - 2 x + 2 + \frac{1}{2} ln x - 1 + x^{2} - 2 x + 2) - \frac{1}{6} ((x^{2} - 2 x + 2)^{\frac{3}{2}} + 3 ln x - 1 + x^{2} - 2 x + 2 (x - 1) - 3 x^{2} - 2 x + 2)

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

Add a constant to the solution

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