JEE Main 202230 Jun, Morning shiftMathematicsMediumIntegral Calculus369× since 2002Let f(t)=∫0tex3(x8(x6+2x3+2)2)dxf(t) = \int\limits_0^t {{e^{{x^3}}}\left( {{{{x^8}} \over {{{({x^6} + 2{x^3} + 2)}^2}}}} \right)dx}f(t)=0∫tex3((x6+2x3+2)2x8)dx. If f(1)+f′(1)=αe−16f(1) + f'(1) = \alpha e - {1 \over 6}f(1)+f′(1)=αe−61, then the value of 150α\alphaα is equal to ___________.Your answerType your answer. Integers and decimals are accepted.CheckSkip · show solution
