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

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

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integral of x^rln(x)

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (arcsin (\frac{y}{x}))

f (x) = \frac{4 x}{x ^{2} + 1}, at x = 0

\int 6 x cos (x) d x

y = x^{2}, y = 3 x

n = 0 \sum \infty 7^{2}

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