How do you integrate $\int x^{4} ln x$ $int x^4lnx$ by integration by parts method?

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Answer 1 · 2016-09-24T23:55:27

$\int x^{n} ln x d x = \frac{1}{n + 1} (x^{n + 1} ln x - \frac{x^{n + 1}}{n + 1}) + C$ $int x^n lnx dx = 1/(n+1)(x^(n+1)lnx-x^(n+1)/(n+1))+C$

$\frac{d}{d x} (x^{n} ln x) = n x^{n - 1} ln x + x^{n - 1}$ $d/(dx)(x^n lnx) = n x^(n-1)lnx+x^(n-1)$ or

$(n + 1) \int x^{n} ln x d x = x^{n + 1} ln x - \int x^{n} d x$ $(n+1)int x^n lnx dx = x^(n+1)lnx-int x^n dx$

so

$\int x^{n} ln x d x = \frac{1}{n + 1} (x^{n + 1} ln x - \frac{x^{n + 1}}{n + 1}) + C$ $int x^n lnx dx = 1/(n+1)(x^(n+1)lnx-x^(n+1)/(n+1))+C$

How do you integrate ∫x4lnxint x^4lnx by integration by parts method?