Question

If \(\int \frac{1}{((x+4)^3 (x+1)^5)^{1/4}} \, dx = A \cdot \left(\frac{x+4}{x+1}\right)^n + c\), then

• A. \(n \cdot A = 3 \)
• B. \(n + \frac{1}{A} = -\frac{1}{2} \)
• C. \(A + n = 1 \)
• D. \(A = n \)

Hint:

For integrals of the form \(\int \frac{1}{(ax+b)^p (cx+d)^q} dx\), a useful substitution is \(t = \frac{ax+b}{cx+d}\). This transforms the integrand into a simpler rational power function of \(t\). Remember to express \(dx\) in terms of \(dt\) and also replace \((ax+b)\) and \((cx+d)\) in terms of \(t\).

Answer 1

The Correct Option is B

Step 1: Rewrite the integrand.
The given integral is \(\int \frac{1}{((x+4)^3 (x+1)^5)^{1/4}} dx\). This can be written as: \[ \int \frac{1}{(x+4)^{3/4} (x+1)^{5/4}} dx \] Step 2: Apply a suitable substitution.
For integrals of the form \(\int \frac{1}{(ax+b)^p (cx+d)^q} dx\), a common substitution is \(t = \frac{ax+b}{cx+d}\).
Let \(t = \frac{x+4}{x+1}\).
Now, find \(dx\) in terms of \(dt\): \[ t = \frac{x+4}{x+1} \implies t(x+1) = x+4 \] \[ tx + t = x + 4 \] \[ tx - x = 4 - t \] \[ x(t - 1) = 4 - t \] \[ x = \frac{4 - t}{t - 1} \] Differentiate \(x\) with respect to \(t\): \[ \frac{dx}{dt} = \frac{-1(t - 1) - (4 - t)(1)}{(t - 1)^2} = \frac{-t + 1 - 4 + t}{(t - 1)^2} = \frac{-3}{(t - 1)^2} \] So, \(dx = \frac{-3}{(t - 1)^2} dt\). Next, express \((x+4)\) and \((x+1)\) in terms of \(t\): \[ x+1 = \frac{4-t}{t-1} + 1 = \frac{4-t + t-1}{t-1} = \frac{3}{t-1} \] \[ x+4 = \frac{4-t}{t-1} + 4 = \frac{4-t + 4t-4}{t-1} = \frac{3t}{t-1} \] Step 3: Substitute into the integral and simplify.
Substitute \(x+4\), \(x+1\), and \(dx\) into the integral:
\[ \int \frac{1}{\left(\frac{3t}{t-1}\right)^{3/4} \left(\frac{3}{t-1}\right)^{5/4}} \cdot \frac{-3}{(t-1)^2} dt \] Simplify the denominator: \[ \left(\frac{3t}{t-1}\right)^{3/4} \left(\frac{3}{t-1}\right)^{5/4} = \frac{3^{3/4} t^{3/4}}{(t-1)^{3/4}} \cdot \frac{3^{5/4}}{(t-1)^{5/4}} \] \[ = \frac{3^{3/4 + 5/4} t^{3/4}}{(t-1)^{3/4 + 5/4}} = \frac{3^{8/4} t^{3/4}}{(t-1)^{8/4}} = \frac{3^2 t^{3/4}}{(t-1)^2} = \frac{9 t^{3/4}}{(t-1)^2} \] Now, substitute this back into the integral: \[ \int \frac{1}{\frac{9 t^{3/4}}{(t-1)^2}} \cdot \frac{-3}{(t-1)^2} dt \] \[ = \int \frac{(t-1)^2}{9 t^{3/4}} \cdot \frac{-3}{(t-1)^2} dt \] Cancel out \((t-1)^2\): \[ = \int \frac{-3}{9 t^{3/4}} dt = \int -\frac{1}{3} t^{-3/4} dt \] Step 4: Perform the integration. \[ -\frac{1}{3} \int t^{-3/4} dt = -\frac{1}{3} \cdot \frac{t^{-3/4 + 1}}{-3/4 + 1} + c \] \[ = -\frac{1}{3} \cdot \frac{t^{1/4}}{1/4} + c = -\frac{1}{3} \cdot 4 t^{1/4} + c \] \[ = -\frac{4}{3} t^{1/4} + c \] Step 5: Substitute back \(t = \frac{x+4}{x+1}\). \[ = -\frac{4}{3} \left(\frac{x+4}{x+1}\right)^{1/4} + c \] Step 6: Compare with the given form and check the options.
The given form is \(A \cdot \left(\frac{x+4}{x+1}\right)^n + c\).
Comparing our result with this form, we find:
\(A = -\frac{4}{3}\)
\(n = \frac{1}{4}\)
Now, let's check the given options: 1. \(n \cdot A = 3 \implies \left(\frac{1}{4}\right) \cdot \left(-\frac{4}{3}\right) = -\frac{1}{3} \neq 3\). (Incorrect)
2. \(n + \frac{1}{A} = -\frac{1}{2} \implies \frac{1}{4} + \frac{1}{(-4/3)} = \frac{1}{4} - \frac{3}{4} = \frac{1 - 3}{4} = \frac{-2}{4} = -\frac{1}{2}\). (Correct)
3. \(A + n = 1 \implies -\frac{4}{3} + \frac{1}{4} = \frac{-16 + 3}{12} = -\frac{13}{12} \neq 1\). (Incorrect)
4. \(A = n \implies -\frac{4}{3} = \frac{1}{4}\). (Incorrect) The correct option is (2). The final answer is $\boxed{n + \frac{1}{A} = -\frac{1}{2}}$.