Key Concepts and Formulas

• Substitution Method for Integration: If $\int f(g(x))g'(x) \, dx$ is difficult to solve directly, substitute $u = g(x)$, then $du = g'(x) \, dx$, which transforms the integral into $\int f(u) \, du$.
• Indefinite Integral: The indefinite integral of a function $f(x)$ is a function $F(x)$ whose derivative is $f(x)$, i.e., $F'(x) = f(x)$. It is denoted by $\int f(x) \, dx = F(x) + C$, where $C$ is the constant of integration.
• Logarithmic Properties: $\ln(a) - \ln(b) = \ln(\frac{a}{b})$, $n \ln(a) = \ln(a^n)$, and $\log_e(x) = \ln(x)$.

Step-by-Step Solution

Step 1: Substitution We are given the integral $\int \frac{1}{x^{1 / 4}\left(1+x^{1 / 4}\right)} \mathrm{d} x$. To simplify this, we use the substitution $x = t^4$. This implies that $dx = 4t^3 \, dt$. WHY: This substitution will eliminate the fractional exponents, making the integral easier to manage.

Step 2: Integral Transformation Substituting $x = t^4$ and $dx = 4t^3 \, dt$ into the integral, we get $\int \frac{1}{x^{1/4}(1 + x^{1/4})} \, dx = \int \frac{1}{t(1 + t)} 4t^3 \, dt = \int \frac{4t^2}{1 + t} \, dt$ WHY: To express the integral in terms of $t$.

Step 3: Rewriting the Integral We can rewrite the integrand $\frac{4t^2}{1 + t}$ using polynomial long division or by adding and subtracting terms in the numerator. $\frac{4t^2}{1+t} = \frac{4(t^2-1+1)}{t+1} = \frac{4(t^2 - 1)}{t+1} + \frac{4}{t+1} = \frac{4(t-1)(t+1)}{t+1} + \frac{4}{t+1} = 4(t-1) + \frac{4}{t+1} = 4t - 4 + \frac{4}{t+1}$ Therefore, $\int \frac{4t^2}{1+t} \, dt = \int \left(4t - 4 + \frac{4}{t+1}\right) \, dt$ WHY: To simplify the integrand into a form that is easier to integrate.

Step 4: Separate and Integrate Now we can integrate each term separately: $\int \left(4t - 4 + \frac{4}{t+1}\right) \, dt = 4 \int t \, dt - 4 \int 1 \, dt + 4 \int \frac{1}{t+1} \, dt$ $= 4 \cdot \frac{t^2}{2} - 4t + 4 \ln|t+1| + C = 2t^2 - 4t + 4 \ln|t+1| + C$ WHY: Applying standard integration rules.

Step 5: Substitute back $t = x^{1/4}$ Substituting $t = x^{1/4}$ back into the expression, we obtain $f(x) = 2(x^{1/4})^2 - 4x^{1/4} + 4 \ln(x^{1/4} + 1) + C = 2x^{1/2} - 4x^{1/4} + 4 \ln(x^{1/4} + 1) + C$ WHY: To express $f(x)$ in terms of $x$.

Step 6: Using the Condition $f(0) = -6$ We are given that $f(0) = -6$. Plugging $x = 0$ into our expression for $f(x)$, we have: $f(0) = 2(0)^{1/2} - 4(0)^{1/4} + 4 \ln(0^{1/4} + 1) + C = 0 - 0 + 4 \ln(1) + C = 0 + C = -6$ Thus, $C = -6$.

WHY: To determine the constant of integration.

Step 7: Finding $f(1)$ Now we have $f(x) = 2x^{1/2} - 4x^{1/4} + 4 \ln(x^{1/4} + 1) - 6$. We want to find $f(1)$: $f(1) = 2(1)^{1/2} - 4(1)^{1/4} + 4 \ln(1^{1/4} + 1) - 6 = 2(1) - 4(1) + 4 \ln(1 + 1) - 6 = 2 - 4 + 4 \ln(2) - 6 = -8 + 4 \ln(2) = 4(\ln(2) - 2)$

WHY: To find the value of the function at the specified point.

Common Mistakes & Tips

• Algebraic Manipulation: Be careful with algebraic manipulation, especially when simplifying the integral after substitution.
• Constant of Integration: Don't forget to add the constant of integration, $C$, when evaluating indefinite integrals. Use the given initial condition to solve for $C$.
• Logarithm Properties: Remember the properties of logarithms to simplify the final answer.

Summary

We used substitution to simplify the integral, then integrated each term to find $f(x)$. Then we used the initial condition $f(0) = -6$ to find the constant of integration. Finally, we calculated $f(1)$ to obtain the final answer.

The final answer is $\boxed{4(\log _{\mathrm{e}} 2-2)}$, which corresponds to option (A).