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

The integral of ln(x^2+x+1) is xln(x^2+x+1)+1/2 ln(x^2+x+1)+sqrt(3)arctan(1/(sqrt(3))(2x+1))-2x-1+C

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Apply Integration By Parts

: u ln (u^{2} + \frac{3}{4}) - \int \frac{8 u ^{2}}{4 u ^{2} + 3} d u

= u ln (u^{2} + \frac{3}{4}) - \int \frac{8 u ^{2}}{4 u ^{2} + 3} d u

\int \frac{8 u ^{2}}{4 u ^{2} + 3} d u = - 3 arctan (\frac{2}{3} u) + 2 u

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

\int 8 sec^{3} (x) d x

\frac{\partial}{\partial y} (x^{2} \cdot ln (y))

4 y^{^{''}} - 6 y^{^{'}} + 7 y = 15

\int 0 cos (x) d x

x (x^{2} - 10 x + 30), x^{2}

What is the integral of ln(x^2+x+1) ?
The integral of ln(x^2+x+1) is xln(x^2+x+1)+1/2 ln(x^2+x+1)+sqrt(3)arctan(1/(sqrt(3))(2x+1))-2x-1+C