\inte^{-3x}(x^2+sin4x)dx=

Correct Answer is : −e−3x(x23+2x9+227+325sin4x+425cos4x)+C

-e^{-3x}({x^2}/3+{2x}/9+2/{27}+3/{25}sin4x+4/{25}cos4x)+C

-e^{-3x}({x^2}/3-{2x}/9+2/{27}+3/{25}sin4x+4/{25}cos4x)+C

-e^{-3x}({x^2}/3+{2x}/9+2/{27}+3/{25}sin4x-4/{25}cos4x)+C

-e^{-3x}({x^2}/3-{2x}/9+2/{27}+3/{25}sin4x-4/{25}cos4x)+C

TS EAMCET 10-Sep-2020 Shift 1 Solved Paper

Section: Mathematics

Question : 73 of 160

Marks: +1, -0

∫e−3x(x2+sin4x)dx=

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