JEE MAIN - Mathematics (2020 - 6th September Evening Slot - No. 5)
The integral $$\int\limits_1^2 {{e^x}.{x^x}\left( {2 + {{\log }_e}x} \right)} dx$$ equals :
e(4e + 1)
e(2e – 1)
e(4e – 1)
4e2 – 1
Explanation
$$\int\limits_1^2 {{e^x}.{x^x}\left( {2 + {{\log }_e}x} \right)} dx$$
= $$\int\limits_1^2 {{e^x}{x^x}\left[ {1 + \left( {1 + {{\log }_e}x} \right)} \right]} dx$$
= $$\int\limits_1^2 {{e^x}\left[ {{x^x} + {x^x}\left( {1 + {{\log }_e}x} \right)} \right]} dx$$
= $$\left[ {{e^x}{x^x}} \right]_1^2$$
= e2 $$ \times $$ 4 - e $$ \times $$ 1
= 4e2 - e
= e(4e - 1)
Note : $$\int {{e^x}\left( {f\left( x \right) + f'\left( x \right)} \right)dx} $$ = exf(x) + c
= $$\int\limits_1^2 {{e^x}{x^x}\left[ {1 + \left( {1 + {{\log }_e}x} \right)} \right]} dx$$
= $$\int\limits_1^2 {{e^x}\left[ {{x^x} + {x^x}\left( {1 + {{\log }_e}x} \right)} \right]} dx$$
= $$\left[ {{e^x}{x^x}} \right]_1^2$$
= e2 $$ \times $$ 4 - e $$ \times $$ 1
= 4e2 - e
= e(4e - 1)
Note : $$\int {{e^x}\left( {f\left( x \right) + f'\left( x \right)} \right)dx} $$ = exf(x) + c
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