The integral $\int\limits_1^2 {{e^x}.{x^x}\left( {2 + {{\log }_e}x} \right)} dx$ equals :
Solution
$\int\limits_1^2 {{e^x}.{x^x}\left( {2 + {{\log }_e}x} \right)} dx$
<br><br>= $$\int\limits_1^2 {{e^x}{x^x}\left[ {1 + \left( {1 + {{\log }_e}x} \right)} \right]} dx$$
<br><br>= $$\int\limits_1^2 {{e^x}\left[ {{x^x} + {x^x}\left( {1 + {{\log }_e}x} \right)} \right]} dx$$
<br><br>= $\left[ {{e^x}{x^x}} \right]_1^2$
<br><br>= e<sup>2</sup> $\times$ 4 - e $\times$ 1
<br><br>= 4e<sup>2</sup> - e
<br><br>= e(4e - 1)
<br><br><b>Note :</b> $\int {{e^x}\left( {f\left( x \right) + f'\left( x \right)} \right)dx}$ = e<sup>x</sup>f(x) + c
About this question
Subject: Mathematics · Chapter: Definite Integration · Topic: Properties of Definite Integrals
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