Let $$f(t) = \int\limits_0^t {{e^{{x^3}}}\left( {{{{x^8}} \over {{{({x^6} + 2{x^3} + 2)}^2}}}} \right)dx} $$. If $f(1) + f'(1) = \alpha e - {1 \over 6}$, then the value of 150$\alpha$ is equal to ___________.

<p>Given,</p> <p>$$f(t) = \int\limits_0^t {{e^{{x^3}}}\left( {{{{x^8}} \over {{{({x^6} + 2{x^3} + 2)}^2}}}} \right)dx} $$</p> <p>$f'(t) = {e^{{t^3}}}\left( {{{{t^8}} \over {{{({t^6} + 2{t^3} + 2)}^2}}}} \right)$</p> <p>$\therefore$ $f'(1) = {e^1}\left( {{1 \over {{{(1 + 2 + 2)}^2}}}} \right)$</p> <p>$= {e \over {{5^2}}}$</p> <p>Now, $$f(t) = \int\limits_0^t {{e^{{x^3}}}\left( {{{{x^8}} \over {{{({x^6} + 2{x^2} + 2)}^2}}}} \right)dx} $$</p> <p>Let ${x^3} = z \Rightarrow 3{x^2}dx = dz$</p> <p>$$ = \int\limits_0^{{t^3}} {{e^z}\left( {{{{x^6}\,.\,{x^2}dx} \over {{{({x^6} + 2{x^3} + 2)}^2}}}} \right)} $$</p> <p>$$ = {1 \over 3}\int\limits_0^{{t^3}} {{e^z}\left( {{{{z^2}dz} \over {{{({z^2} + 2z + 2)}^2}}}} \right)} $$</p> <p>$$ = {1 \over 3}\int\limits_0^{{t^3}} {{e^z}\left( {{{{z^2} + 2z + 2 - 2z - 2} \over {{{({z^2} + 2z + 2)}^2}}}} \right)} dz$$</p> <p>$$ = {1 \over 3}\int\limits_0^{{t^3}} {{e^z}\left( {{{{z^2} + 2z + 2} \over {{{({z^2} + 2z + 2)}^2}}} - {{2z + 2} \over {{{({z^2} + 2z + 2)}^2}}}} \right)dz} $$</p> <p>$$ = {1 \over 3}\int\limits_0^{{t^3}} {{e^z}\left( {{1 \over {{z^2} + 2z + 2}} - {{2z + 2} \over {{{({z^2} + 2z + 2)}^2}}}} \right)dz} $$</p> <p>$= {1 \over 3}\int\limits_0^{{t^3}} {{e^z}\left( {f(z) + f'(z)} \right)dz}$</p> <p>$= {1 \over 3}\left[ {{e^z}f(z)} \right]_0^{{t^3}}$</p> <p>$= {1 \over 3}\left[ {{e^z} \times {1 \over {{z^2} + 2z + 2}}} \right]_0^{{t^3}}$</p> <p>$$ = {1 \over 3}\left[ {{e^{{t^3}}} \times {1 \over {{t^6} + 2{t^3} + 2}} - {1 \over 2}} \right]$$</p> <p>$\therefore$ $$f(t) = {1 \over 3}\left[ {{{{e^{{t^3}}}} \over {{t^6} + 2{t^3} + 2}} - {1 \over 2}} \right]$$</p> <p>So, $f(1) = {1 \over 3}\left[ {{e \over {1 + 2 + 2}} - {1 \over 2}} \right]$</p> <p>$= {1 \over 3}\left[ {{e \over 5} - {1 \over 2}} \right]$</p> <p>Given, $f(1) + f'(1) = \alpha e - {1 \over 6}$</p> <p>$$ \Rightarrow {1 \over 3}\left[ {{e \over 5} - {1 \over 2}} \right] + {e \over {25}} = \alpha e - {1 \over 6}$$</p> <p>$$ \Rightarrow {e \over {15}} - {1 \over 6} + {e \over {25}} = \alpha e - {1 \over 6}$$</p> <p>$\Rightarrow {e \over {15}} + {e \over {25}} = \alpha e$</p> <p>$\Rightarrow {{10e + 6e} \over {150}} = \alpha e$</p> <p>$\Rightarrow {{16e} \over {150}} = \alpha e$</p> <p>$\Rightarrow \alpha = {{16} \over {150}}$</p> <p>$\therefore$ $150\alpha = 150 \times {{16} \over {150}} = 16$</p>