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Step-by-step Solution
Step 1: Define the function f(t)
We have
f(t) = \int_{0}^{t} e^{x^3} \left(\frac{x^8}{\bigl(x^6 + 2x^3 + 2\bigr)^2}\right)\,dx.
Step 2: Find f'(t)
By the Fundamental Theorem of Calculus, the derivative of a definite integral from 0 to t is simply the integrand evaluated at x = t:
f'(t) = e^{t^3} \left(\frac{t^8}{\bigl(t^6 + 2t^3 + 2\bigr)^2}\right).
Step 3: Evaluate f'(1)
Substituting t = 1 into the expression for f'(t):
f'(1) = e^{1^3} \left(\frac{1^8}{\bigl(1^6 + 2\cdot1^3 + 2\bigr)^2}\right)
= e \,\frac{1}{5^2}
= \frac{e}{25}.
Step 4: Attempt to integrate f(t) using a substitution
To find an exact expression for
\int_{0}^{t} e^{x^3} \left(\frac{x^8}{\bigl(x^6 + 2x^3 + 2\bigr)^2}\right) dx,
we make the substitution:
z = x^3 \quad\Longrightarrow\quad 3x^2\,dx = dz \quad\text{or}\quad x^2\,dx = \frac{dz}{3}.
Note that
x^6 = (x^3)^2 = z^2, \quad x^8 = x^6 \cdot x^2 = z^2 \cdot x^2.
Hence,
x^8 \,dx = z^2 \,x^2 \,dx = z^2 \cdot \frac{dz}{3}.
Also,
x^6 + 2x^3 + 2 = z^2 + 2z + 2.
So the integral becomes
f(t)
= \int_{0}^{t} e^{x^3}\,\frac{x^8}{\bigl(x^6 + 2x^3 + 2\bigr)^2}\,dx
= \int_{0}^{t^3} e^{z}\,\frac{z^2}{\bigl(z^2 + 2z + 2\bigr)^2}\,\frac{dz}{3}.
Thus,
f(t) = \frac{1}{3}\,\int_{0}^{t^3} e^z\,\frac{z^2}{(z^2 + 2z + 2)^2}\,dz.
Step 5: Rewrite the integrand in a form suitable for a known derivative
Observe that we want to utilize a form that looks like e^z(f(z) + f'(z)) structure if possible. Notice:
\frac{z^2}{(z^2 + 2z + 2)^2}
= \frac{z^2 + 2z + 2 - (2z + 2)}{(z^2 + 2z + 2)^2}
= \frac{1}{z^2 + 2z + 2} - \frac{2z + 2}{(z^2 + 2z + 2)^2}.
If we set
f(z) = \frac{1}{z^2 + 2z + 2},
then
f'(z) = -\,\frac{2z + 2}{(z^2 + 2z + 2)^2}.
Thus,
\frac{z^2}{(z^2 + 2z + 2)^2}
= f(z) + f'(z).
So the integral becomes:
f(t)
= \frac{1}{3}\,\int_{0}^{t^3} e^z\Bigl(f(z) + f'(z)\Bigr)\,dz.
Step 6: Use the known result: ∫ e^z (f(z) + f'(z)) dz = e^z f(z)
We know that
\int e^z \bigl(f(z) + f'(z)\bigr)\,dz = e^z f(z) + C.
Hence,
f(t)
= \frac{1}{3}\left[\bigl(e^z f(z)\bigr)\Big|_{0}^{t^3}\right]
= \frac{1}{3}\left[e^{t^3}\,\frac{1}{t^6 + 2t^3 + 2} \;-\; e^{0}\,\frac{1}{0^2 + 2\cdot0 + 2}\right].
Since e^0 = 1 and (0^2 + 2\cdot0 + 2) = 2 , we get
f(t)
= \frac{1}{3}\left[\frac{e^{t^3}}{t^6 + 2t^3 + 2} - \frac{1}{2}\right].
Step 7: Evaluate f(1)
Substituting t = 1 ,
f(1)
= \frac{1}{3}\left[\frac{e^{1^3}}{1^6 + 2\cdot1^3 + 2} - \frac{1}{2}\right]
= \frac{1}{3}\left[\frac{e}{5} - \frac{1}{2}\right].
Step 8: Compute f(1) + f'(1)
We already found
f'(1) = \frac{e}{25}.
Thus,
f(1) + f'(1)
= \frac{1}{3}\left[\frac{e}{5} - \frac{1}{2}\right] + \frac{e}{25}.
We are given that
f(1) + f'(1) = \alpha e - \frac{1}{6}.
Step 9: Equate terms to find α
Rewrite:
\frac{1}{3}\left[\frac{e}{5} - \frac{1}{2}\right] + \frac{e}{25}
= \alpha e - \frac{1}{6}.
Simplify the left side:
The term with e:
\frac{1}{3} \cdot \frac{e}{5} = \frac{e}{15},
\quad
\text{and}
\quad
\frac{e}{25}.
Hence, total e-term is
\frac{e}{15} + \frac{e}{25} = \frac{5e + 3e}{75} = \frac{8e}{75},
but carefully re-checking the common denominator of 15 and 25 is 75 indeed.
Alternatively:
\frac{e}{15} = \frac{5e}{75}, \quad \frac{e}{25} = \frac{3e}{75} \quad\Longrightarrow\quad \frac{8e}{75}.
The constant term:
\frac{1}{3}\left(-\,\frac{1}{2}\right) = -\frac{1}{6}.
So the left side is
\frac{8e}{75} - \frac{1}{6}.
Thus,
\frac{8e}{75} - \frac{1}{6}
= \alpha e - \frac{1}{6}.
Canceling the like constant on both sides (−1/6), we get
\frac{8e}{75} = \alpha e.
Hence,
\alpha = \frac{8}{75}.
Step 10: Find 150α
150\alpha = 150 \times \frac{8}{75} = 2 \times 8 = 16.
Final Answer
The value of 150\alpha is 16.
