The integral `I = int (dx)/((x+1)^(3//4)(x-2)^(5//4))` is equal to

Options

A. `4((x+1)/(x-2))^(1//4)`
B. `4((x-2)/(x+2))^(1//2)`
C. `-(4)/(3)((x+1)/(x-2))^(1//4)`
D. `-(4)/(3)((x-2)/(x+1))^(1//4)`

Answer

To solve the integral \( I = \int \frac{dx}{(x+1)^{3/4}(x-2)^{5/4}} \), we will use substitution and integration techniques.

### Step-by-Step Solution:

1. **Substitution**:
Let \( t = (x + 1)^{1/4} \). Then, we have:
\[ x + 1 = t^4 \implies x = t^4 - 1 \]
Differentiating both sides gives:
\[ dx = 4t^3 dt \]

2. **Rewrite the Integral**:
Substitute \( x \) and \( dx \) into the integral:
\[ I = \int \frac{4t^3 dt}{(t^4)^{3/4}((t^4 - 1) - 2)^{5/4}} = \int \frac{4t^3 dt}{t^3(t^4 - 3)^{5/4}} \]
Simplifying gives:
\[ I = 4 \int \frac{dt}{(t^4 - 3)^{5/4}} \]

3. **Further Simplification**:
To simplify the integral, we can multiply the numerator and denominator by \( t^2 \):
\[ I = 4 \int \frac{t^2 dt}{t^2(t^4 - 3)^{5/4}} = 4 \int \frac{dt}{(t^4 - 3)^{5/4} t^2} \]

4. **Substitution for Integration**:
Let \( u = t^4 - 3 \). Then, we have:
\[ du = 4t^3 dt \implies dt = \frac{du}{4t^3} \]
Substitute \( t^3 = (u + 3)^{3/4} \):
\[ I = 4 \int \frac{(u + 3)^{3/4}}{u^{5/4}} \frac{du}{4(u + 3)^{3/4}} = \int \frac{du}{u^{5/4}} \]

5. **Integrate**:
Now we can integrate:
\[ I = \int u^{-5/4} du = \frac{u^{-1/4}}{-1/4} + C = -4u^{-1/4} + C \]

6. **Back Substitute**:
Substitute back \( u = t^4 - 3 \):
\[ I = -4(t^4 - 3)^{-1/4} + C \]
Now substitute back \( t = (x + 1)^{1/4} \):
\[ I = -4((x + 1) - 3)^{-1/4} + C = -4((x - 2)^{1/4})^{-1} + C \]

7. **Final Result**:
Thus, the final answer is:
\[ I = -4(x - 2)^{-1/4} + C \]