What is the integral of (4x^3-3x^2+1)ln(x) ?

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

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integral of (4x^3-3x^2+1)ln(x)

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

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

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

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