$\int x^{x} log (e x) d x$ is equal to

x^{x} + c

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

x \cdot log x + c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{(log x)}^{x} + c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{log x} + c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $x^{x} + c$
Let $I = \int x^{x} (log e x) d x$
$= \int x^{x} (1 + log x) d x$
Let $t = x^{x} = e^{x log x}$
$\Rightarrow \frac{d t}{d x} = x^{x} {x \cdot \frac{1}{x} + log x}$
$\Rightarrow d t = x^{x} (1 + log x) d x$
Therefore, $I = \int t + c = x^{x} + c$

Suggest Corrections

Similar questions

(log x)^{x} + x^{log x}

x

\frac{d y}{d x}

y = x^{\log x} + {(\log x)}^{x}

{log}_{x} x - {log}_{x} y = a, {log}_{x} y - {log}_{x} z = b & {log}_{x} z - {log}_{x} x = C

{(\frac{x}{y})}^{b - c} \times {(\frac{y}{z})}^{c - a} \times {(\frac{z}{x})}^{a - b}

x^{log x} + {(log x)}^{x}

x

(log x)^{x} + x^{log x}

Join BYJU'S Learning Program