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

The integral of (x^3+1)ln(x) is 1/4 x^4ln(x)+xln(x)-(x^4)/(16)-x+C

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integral of (x^3+1)ln(x)

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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