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

The integral of (x^3-2x+3)ln(x) is 1/4 x^4ln(x)-(x^4}{16}-x^2ln(x)+\frac{x^2)/2+3xln(x)-3x+C

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

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

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

Solution steps

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

Expand

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int x^{3} ln (x) d x = \frac{1}{4} x^{4} ln (x) - \frac{x ^{4}}{16}

\int 2 x ln (x) d x = x^{2} ln (x) - \frac{x ^{2}}{2}

\int 3 ln (x) d x = 3 (x ln (x) - x)

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

Simplify

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

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

Add a constant to the solution

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

What is the integral of (x^3-2x+3)ln(x) ?
The integral of (x^3-2x+3)ln(x) is 1/4 x^4ln(x)-(x^4}{16}-x^2ln(x)+\frac{x^2)/2+3xln(x)-3x+C