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How do you integrate $(x^{4}) (ln x)$ $(x^4)(lnx)$ ?

1 Answer

Shwetank Mauria

Jul 26, 2016

$\int ln x \times x^{4} d x = \frac{x^{5}}{5} (ln x - \frac{1}{5})$ $intlnx xxx^4dx=x^5/5(lnx-1/5)$

Explanation:

WE can use integration by parts $\int u d v = u v - \int v d u$ $intudv=uv-intvdu$

Let $u = ln x$ $u=lnx$ and $v = \frac{x^{5}}{5}$ $v=x^5/5$

Hence $d u = \frac{d x}{x}$ $du=dx/x$ and $d v = x^{4} d x$ $dv=x^4dx$ and $\int u d v = u v = \int v d u$ $intudv=uv=intvdu$ is

$\int ln x \times x^{4} d x = \int u d v = u v - \int v d u$ $intlnx xxx^4dx=intudv=uv-intvdu$

= $\frac{x^{5}}{5} \times ln x - \int \frac{x^{5}}{5} \times \frac{d x}{x}$ $x^5/5xxlnx-intx^5/5xxdx/x$

= $\frac{ln x \times x^{5}}{5} - \frac{1}{5} \int x^{4} d x$ $(lnx xx x^5)/5-1/5intx^4dx$

= $\frac{ln x \times x^{5}}{5} - \frac{x^{5}}{25}$ $(lnx xx x^5)/5-x^5/25$

= $\frac{x^{5}}{5} (ln x - \frac{1}{5})$ $x^5/5(lnx-1/5)$

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