`int (dx)/(root3(x)+root4(x))`

Answer

To solve the integral \(\int \frac{dx}{\sqrt[3]{x} + \sqrt[4]{x}}\), we can follow these steps:

### Step 1: Rewrite the integral
We start by rewriting the integral in terms of exponents:
\[ I = \int \frac{dx}{x^{1/3} + x^{1/4}} \]

### Step 2: Factor out the lower power
Next, we notice that the lower power between \(x^{1/3}\) and \(x^{1/4}\) is \(x^{1/4}\). We can factor \(x^{1/4}\) out of the denominator:
\[ I = \int \frac{dx}{x^{1/4} \left(x^{1/3 - 1/4} + 1\right)} = \int \frac{dx}{x^{1/4} \left(x^{1/12} + 1\right)} \]

### Step 3: Substitute for simplification
Let \(t = x^{1/12} + 1\). Then, we differentiate both sides to find \(dx\):
\[ dt = \frac{1}{12} x^{-11/12} dx \quad \Rightarrow \quad dx = 12 x^{11/12} dt \]
Now, we express \(x^{11/12}\) in terms of \(t\):
\[ x^{11/12} = (t - 1)^{11/12} \]
Substituting \(dx\) into the integral:
\[ I = \int \frac{12 (t - 1)^{11/12} dt}{(t - 1)^{3/4} (t)} = 12 \int \frac{(t - 1)^{2/3}}{t} dt \]

### Step 4: Expand and simplify the integral
Now we can expand the integrand:
\[ I = 12 \int \left( (t - 1)^{2/3} \cdot \frac{1}{t} \right) dt \]
This can be computed using polynomial long division or binomial expansion.

### Step 5: Integrate term by term
Using the binomial expansion, we can express \((t - 1)^{2/3}\) and integrate term by term:
\[ I = 12 \left[ \int t^{5/3} dt - \int t^{2/3} dt \right] \]

### Step 6: Solve the integrals
Now we compute each integral:
\[ \int t^{5/3} dt = \frac{t^{8/3}}{8/3} = \frac{3}{8} t^{8/3} \]
\[ \int t^{2/3} dt = \frac{t^{5/3}}{5/3} = \frac{3}{5} t^{5/3} \]
Putting it all together:
\[ I = 12 \left( \frac{3}{8} t^{8/3} - \frac{3}{5} t^{5/3} \right) + C \]

### Step 7: Substitute back for \(t\)
Finally, we substitute back \(t = x^{1/12} + 1\):
\[ I = 12 \left( \frac{3}{8} (x^{1/12} + 1)^{8/3} - \frac{3}{5} (x^{1/12} + 1)^{5/3} \right) + C \]

### Final Answer
Thus, the integral evaluates to:
\[ I = \frac{9}{2} (x^{1/12} + 1)^{8/3} - \frac{36}{5} (x^{1/12} + 1)^{5/3} + C \]