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

The integral of (1+x+x^2)ln(x) is xln(x)+1/2 x^2ln(x)+1/3 x^3ln(x)-x-(x^2)/4-(x^3)/9+C

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{1}{6} (6 + 3 x + 2 x^{2}) d x = x + \frac{x ^{2}}{4} + \frac{x ^{3}}{9}

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

Simplify

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

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

Add a constant to the solution

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

y = - csc (x) - sin (x)

y^{^{''}} - 4 y^{^{'}} - 5 y = - 3 e^{4} t, y (0) = 0, y^{^{'}} (0) = 0

\int_{1}^{x} (\frac{1}{t ^{3} + 1}) d t

\int \frac{1}{sec ( a x ) + 1} d x

\frac{d y}{d x} = \frac{( ( 4 x ) sec ( \frac{y}{x} ) + y )}{x}

What is the integral of (1+x+x^2)ln(x) ?
The integral of (1+x+x^2)ln(x) is xln(x)+1/2 x^2ln(x)+1/3 x^3ln(x)-x-(x^2)/4-(x^3)/9+C