Evaluate: ` int ( dx)/( sqrt(x+1) +""^(3) sqrt(x+1))`

Answer

To evaluate the integral

\[ \int \frac{dx}{\sqrt{x+1} + (x+1)^{\frac{1}{3}}} \]

we will follow a systematic approach using substitution.

### Step 1: Rewrite the Integral
First, we rewrite the integral for clarity:

\[ \int \frac{dx}{(x+1)^{\frac{1}{2}} + (x+1)^{\frac{1}{3}}} \]

### Step 2: Substitution
We will use the substitution \( t = (x + 1)^{\frac{1}{6}} \). This means \( x + 1 = t^6 \) and thus \( x = t^6 - 1 \).

Now, we differentiate \( x + 1 \):

\[ dx = 6t^5 dt \]

### Step 3: Substitute in the Integral
Now we substitute \( x + 1 \) and \( dx \) in the integral:

\[ \int \frac{6t^5 dt}{t^3 + t^2} \]

### Step 4: Simplify the Integral
We can factor out \( t^2 \) from the denominator:

\[ = \int \frac{6t^5 dt}{t^2(t + 1)} = 6 \int \frac{t^3 dt}{t + 1} \]

### Step 5: Split the Integral
Now we can split the integral into two parts:

\[ = 6 \int \left(t^2 - 1 + \frac{1}{t + 1}\right) dt \]

### Step 6: Integrate Each Term
Now we integrate each term separately:

1. \( 6 \int t^2 dt = 6 \cdot \frac{t^3}{3} = 2t^3 \)
2. \( 6 \int -1 dt = -6t \)
3. \( 6 \int \frac{1}{t + 1} dt = 6 \log(t + 1) \)

Combining these results, we have:

\[ 2t^3 - 6t + 6 \log(t + 1) + C \]

### Step 7: Substitute Back
Now we substitute back \( t = (x + 1)^{\frac{1}{6}} \):

\[ = 2((x + 1)^{\frac{1}{6}})^3 - 6((x + 1)^{\frac{1}{6}}) + 6 \log((x + 1)^{\frac{1}{6}} + 1) + C \]

This simplifies to:

\[ = 2(x + 1)^{\frac{1}{2}} - 6(x + 1)^{\frac{1}{6}} + 6 \log((x + 1)^{\frac{1}{6}} + 1) + C \]

### Final Answer
Thus, the evaluated integral is:

\[ 2(x + 1)^{\frac{1}{2}} - 6(x + 1)^{\frac{1}{6}} + 6 \log((x + 1)^{\frac{1}{6}} + 1) + C \]