Question

\(∫\frac{dx}{(x-1)^{34} (x+2)^{\frac54}}=\)

• A. \(\frac{4}{3}(\frac{x-1}{x+2})^{\frac14}+c\)
• B. \(\frac{3}{4}(\frac{x-1}{x-2})^{\frac14}+c\)
• C. \(\frac{4}{3}(\frac{x+2}{x-1})^{\frac14}+c\)
• D. \(\frac{4}{3}(\frac{x-1=2}{x-1})^{\frac14}+c\)

Answer 1

The Correct Option is A

To solve the problem, we need to evaluate the integral $I = \int \frac{dx}{(x-1)^{3/4} (x+2)^{5/4}}$.

1. Rewrite the Integrand:
Rewrite the denominator:
$(x+2)^{5/4} = (x+2)^{3/4} (x+2)^{2/4} = (x+2)^{3/4} (x+2)^{1/2}$.
Thus:
$I = \int \frac{dx}{(x-1)^{3/4} (x+2)^{3/4} (x+2)^{1/2}} = \int \frac{dx}{\left( \frac{x-1}{x+2} \right)^{3/4} (x+2)^2}$.

2. Substitution:
Let $t = \left( \frac{x-1}{x+2} \right)^{1/4}$, so $t^4 = \frac{x-1}{x+2}$.
Solve for $x$:
$t^4 (x + 2) = x - 1$, so $t^4 x + 2t^4 = x - 1$, and $x (t^4 - 1) = -1 - 2t^4$.
Thus, $x = \frac{1 + 2t^4}{1 - t^4}$.

3. Compute the Differential:
Differentiate $x = \frac{1 + 2t^4}{1 - t^4}$:
$dx = \frac{(1 - t^4)(8t^3) - (1 + 2t^4)(-4t^3)}{(1 - t^4)^2} dt = \frac{8t^3 - 8t^7 + 4t^3 + 8t^7}{(1 - t^4)^2} dt = \frac{12 t^3}{(1 - t^4)^2} dt$.

4. Express $x + 2$ in Terms of $t$:
$x + 2 = \frac{1 + 2t^4}{1 - t^4} + 2 = \frac{1 + 2t^4 + 2 - 2t^4}{1 - t^4} = \frac{3}{1 - t^4}$.

5. Substitute and Simplify:
Substitute into the integral:
$I = \int \frac{\frac{12 t^3}{(1 - t^4)^2}}{t^3 \left( \frac{3}{1 - t^4} \right)^2} dt = \int \frac{12 t^3}{(1 - t^4)^2} \cdot \frac{(1 - t^4)^2}{t^3 \cdot 9} dt = \int \frac{12}{9} dt$.
Integrate:
$\frac{12}{9} \int dt = \frac{4}{3} t + C = \frac{4}{3} \left( \frac{x-1}{x+2} \right)^{1/4} + C$.

Final Answer:
The integral is $\frac{4}{3} \left( \frac{x-1}{x+2} \right)^{1/4} + C$.