What is the integral of (x^2)/(sqrt(x^2-x+1)) ?

The integral of (x^2)/(sqrt(x^2-x+1)) is

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integral of (x^2)/(sqrt(x^2-x+1))

\int \frac{x ^{2}}{x ^{2} - x + 1} d x

\frac{1}{16} (8 x x^{2} - x + 1 + 12 x^{2} - x + 1 - 6 ln 2 x + 3 \frac{4 x ^{2} - 4 x + 4}{3} - 1 + 4 ln \frac{2 x}{3} + \frac{4 x ^{2} - 4 x + 4}{3} - \frac{1}{3} + 3 ln (3)) + C

Solution steps

\int \frac{x ^{2}}{x ^{2} - x + 1} d x

Complete the square

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

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

Apply u-substitution

: \int \frac{( 2 u + 1 ) ^{2}}{2 4 u ^{2} + 3} d u

= \int \frac{( 2 u + 1 ) ^{2}}{2 4 u ^{2} + 3} 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{( 2 u + 1 ) ^{2}}{4 u ^{2} + 3} d u

Expand

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

= \frac{1}{2} \cdot \int \frac{4 u ^{2}}{4 u ^{2} + 3} + \frac{4 u}{4 u ^{2} + 3} + \frac{1}{4 u ^{2} + 3} d u

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} (\int \frac{4 u ^{2}}{4 u ^{2} + 3} d u + \int \frac{4 u}{4 u ^{2} + 3} d u + \int \frac{1}{4 u ^{2} + 3} d u)

\int \frac{4 u ^{2}}{4 u ^{2} + 3} d u = \frac{1}{4} (2 u 4 u^{2} + 3 - 3 (ln 2 u + 3 \frac{4 u ^{2} + 3}{3} - \frac{ln ( 3 )}{2}))

\int \frac{4 u}{4 u ^{2} + 3} d u = 4 u^{2} + 3

\int \frac{1}{4 u ^{2} + 3} d u = \frac{1}{2} ln \frac{2}{3} u + \frac{1}{3} (3 + 4 u^{2})

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

Substitute back

u = x - \frac{1}{2}

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

Simplify

\frac{1}{2} \frac{1}{4} 2 (x - \frac{1}{2}) 4 (x - \frac{1}{2})^{2} + 3 - 3 ln 2 (x - \frac{1}{2}) + 3 \frac{4 ( x - \frac{1}{2} ) ^{2} + 3}{3} - \frac{ln ( 3 )}{2} + 4 (x - \frac{1}{2})^{2} + 3 + \frac{1}{2} ln \frac{2}{3} (x - \frac{1}{2}) + \frac{1}{3} (3 + 4 (x - \frac{1}{2})^{2}) : \frac{1}{16} (8 x x^{2} - x + 1 + 12 x^{2} - x + 1 - 6 ln 2 x + 3 \frac{4 x ^{2} - 4 x + 4}{3} - 1 + 4 ln \frac{2 x}{3} + \frac{4 x ^{2} - 4 x + 4}{3} - \frac{1}{3} + 3 ln (3))

= \frac{1}{16} (8 x x^{2} - x + 1 + 12 x^{2} - x + 1 - 6 ln 2 x + 3 \frac{4 x ^{2} - 4 x + 4}{3} - 1 + 4 ln \frac{2 x}{3} + \frac{4 x ^{2} - 4 x + 4}{3} - \frac{1}{3} + 3 ln (3))

Add a constant to the solution

= \frac{1}{16} (8 x x^{2} - x + 1 + 12 x^{2} - x + 1 - 6 ln 2 x + 3 \frac{4 x ^{2} - 4 x + 4}{3} - 1 + 4 ln \frac{2 x}{3} + \frac{4 x ^{2} - 4 x + 4}{3} - \frac{1}{3} + 3 ln (3)) + C

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What is the integral of (x^2)/(sqrt(x^2-x+1)) ?
The integral of (x^2)/(sqrt(x^2-x+1)) is