Audrey Arnold

2022-11-11

Find Difficult Indefinite Integral $\int \left(\mathrm{log}x+1\right){x}^{x}\phantom{\rule{thinmathspace}{0ex}}dx$

Lillianna Salazar

Expert

The form of the integrand suggests writing
$\left(1+\mathrm{log}x\right){x}^{x}=\left(1+\mathrm{log}x\right){e}^{x\mathrm{log}x},$
then observing that by the product rule,
$\frac{d}{dx}\left[x\mathrm{log}x\right]=x\cdot \frac{1}{x}+1\cdot \mathrm{log}x=1+\mathrm{log}x.$
Consequently, the integrand is of the form ${f}^{\prime }\left(x\right){e}^{f\left(x\right)}$, and its antiderivative is simply
${e}^{f\left(x\right)}={e}^{x\mathrm{log}x}={x}^{x}.$

Expert

Use the substitution $y={x}^{x}$, then do logarithmic differentitation. to get $\left(1+\mathrm{ln}x\right){x}^{x}=\frac{dy}{dx}$