To evaluate the integral
\[
I = \int \frac{x^{1/2}}{1 + x^{3/4}} \, dx,
\]
we will use a substitution method. Let's follow the steps:
### Step 1: Substitute \( x = t^4 \)
We start by letting \( x = t^4 \). Then, we differentiate both sides to find \( dx \):
\[
dx = 4t^3 \, dt.
\]
### Step 2: Rewrite the integral in terms of \( t \)
Now, we can rewrite the integral \( I \):
\[
I = \int \frac{(t^4)^{1/2}}{1 + (t^4)^{3/4}} \cdot 4t^3 \, dt.
\]
This simplifies to:
\[
I = \int \frac{t^2}{1 + t^3} \cdot 4t^3 \, dt = 4 \int \frac{t^5}{1 + t^3} \, dt.
\]
### Step 3: Simplify the integrand
Next, we can simplify the integrand:
\[
I = 4 \int \frac{t^5}{1 + t^3} \, dt.
\]
We can divide \( t^5 \) by \( 1 + t^3 \):
\[
t^5 = t^3(1 + t^3) - t^3,
\]
which gives us:
\[
\frac{t^5}{1 + t^3} = t^3 - \frac{t^3}{1 + t^3}.
\]
### Step 4: Split the integral
Now we can split the integral:
\[
I = 4 \left( \int t^3 \, dt - \int \frac{t^3}{1 + t^3} \, dt \right).
\]
### Step 5: Evaluate the first integral
The first integral is straightforward:
\[
\int t^3 \, dt = \frac{t^4}{4}.
\]
### Step 6: Evaluate the second integral
For the second integral, we can use the substitution \( u = 1 + t^3 \), which gives us \( du = 3t^2 \, dt \) or \( dt = \frac{du}{3t^2} \). We also have \( t^2 = (u - 1)^{2/3} \).
Thus, we can rewrite the integral as:
\[
\int \frac{t^3}{1 + t^3} \, dt = \int \frac{t^3}{u} \cdot \frac{du}{3t^2} = \frac{1}{3} \int \frac{t}{u} \, du.
\]
### Step 7: Combine results
Now we combine the results:
\[
I = 4 \left( \frac{t^4}{4} - \frac{1}{3} \int \frac{t}{u} \, du \right).
\]
### Step 8: Substitute back to \( x \)
Finally, we substitute back \( t = x^{1/4} \) to express the integral in terms of \( x \).
### Final Result
After evaluating the integrals and simplifying, we arrive at the final expression for the integral:
\[
I = \frac{4}{3} \left( x^{3/4} - \log(1 + x^{3/4}) \right) + C,
\]
where \( C \) is the constant of integration.
