What is the integral of xsqrt(x^2+x+1) ?

The integral of xsqrt(x^2+x+1) is

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integral of xsqrt(x^2+x+1)

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

\frac{1}{48} (16 (x^{2} + x + 1)^{\frac{3}{2}} - 3 (4 (x + \frac{1}{2}) x^{2} + x + 1 + 3 (ln 2 x + 1 + 3 \frac{4 x ^{2} + 4 x + 4}{3} - \frac{ln ( 3 )}{2}))) + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Expand

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

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

\int 2 u 4 u^{2} + 3 d u = \frac{1}{6} (4 u^{2} + 3)^{\frac{3}{2}}

\int 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}))

= \frac{1}{4} (\frac{1}{6} (4 u^{2} + 3)^{\frac{3}{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})))

Substitute back

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

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

Simplify

\frac{1}{4} \frac{1}{6} (4 (x + \frac{1}{2})^{2} + 3)^{\frac{3}{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} : \frac{1}{48} (16 (x^{2} + x + 1)^{\frac{3}{2}} - 3 (4 (x + \frac{1}{2}) x^{2} + x + 1 + 3 (ln 2 x + 1 + 3 \frac{4 x ^{2} + 4 x + 4}{3} - \frac{ln ( 3 )}{2})))

= \frac{1}{48} (16 (x^{2} + x + 1)^{\frac{3}{2}} - 3 (4 (x + \frac{1}{2}) x^{2} + x + 1 + 3 (ln 2 x + 1 + 3 \frac{4 x ^{2} + 4 x + 4}{3} - \frac{ln ( 3 )}{2})))

Add a constant to the solution

= \frac{1}{48} (16 (x^{2} + x + 1)^{\frac{3}{2}} - 3 (4 (x + \frac{1}{2}) x^{2} + x + 1 + 3 (ln 2 x + 1 + 3 \frac{4 x ^{2} + 4 x + 4}{3} - \frac{ln ( 3 )}{2}))) + C

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