What is the integral of x^pln(x) ?

The integral of x^pln(x) is 1/(p+1)x^{p+1}ln(x)-(x^{p+1})/((p+1)^2)+C

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integral of x^pln(x)

\int x^{p} ln (x) d x

\frac{1}{p + 1} x^{p + 1} ln (x) - \frac{x ^{p + 1}}{( p + 1 ) ^{2}} + C

Solution steps

\int x^{p} ln (x) d x

Apply Integration By Parts

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

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

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

= \frac{1}{p + 1} x^{p + 1} ln (x) - \frac{x ^{p + 1}}{( p + 1 ) ^{2}}

Add a constant to the solution

= \frac{1}{p + 1} x^{p + 1} ln (x) - \frac{x ^{p + 1}}{( p + 1 ) ^{2}} + C

f (x) = (x^{2} + 9)^{\frac{1}{2}}

5^{y}

\int (r^{3} + 1)^{2} r^{2} d r

\frac{1 - ln ( x )}{x ^{2}}

\int_{4}^{15} 6 x^{- 2} d x

What is the integral of x^pln(x) ?
The integral of x^pln(x) is 1/(p+1)x^{p+1}ln(x)-(x^{p+1})/((p+1)^2)+C