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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{- x}{1 - x} d x

\int sec^{4} (d) x d x

\frac{d h}{d t} = 4 - 16 t^{2}

g (x) = x^{2} + 3 x^{3}

\frac{d y}{d x} (x y) = 7

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