What is the integral of (3x^2+2x)ln(x) ?

The integral of (3x^2+2x)ln(x) is x^3ln(x)+x^2ln(x)-(x^3)/3-(x^2)/2+C

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integral of (3x^2+2x)ln(x)

\int (3 x^{2} + 2 x) ln (x) d x

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

Solution steps

\int (3 x^{2} + 2 x) ln (x) d x

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

(x^{2} + y^{2} + 1) d x + x (x - 2 y) d y = 0

\frac{d}{d x} ((1 + sin (3 x))^{4})

\frac{d}{d x} ((3 cos (2 x) + sin (3 x) + 4)^{2})

y^{^{'}} - 4 y = 10 sin (2 t), y (0) = 0

f (x) = - \frac{3}{e ^{3 x}}

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