Integrate with respect to `x:sqrt(x+1)`

Options

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

Answer

To solve the integral of \(\sqrt{x + 1}\) with respect to \(x\), we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \sqrt{x + 1} \, dx \]

### Step 2: Use Substitution
Let \(t = x + 1\). Then, differentiate both sides with respect to \(x\):
\[ dt = dx \]
This means we can replace \(dx\) in our integral with \(dt\).

### Step 3: Change the Integral
Now, substituting \(t\) into the integral gives:
\[ I = \int \sqrt{t} \, dt \]

### Step 4: Apply the Power Rule for Integration
Recall that \(\sqrt{t} = t^{1/2}\). We can now use the power rule for integration, which states:
\[ \int t^n \, dt = \frac{t^{n + 1}}{n + 1} + C \]
For our case, \(n = \frac{1}{2}\):
\[ I = \int t^{1/2} \, dt = \frac{t^{1/2 + 1}}{1/2 + 1} + C = \frac{t^{3/2}}{3/2} + C = \frac{2}{3} t^{3/2} + C \]

### Step 5: Substitute Back
Now, we substitute back \(t = x + 1\):
\[ I = \frac{2}{3} (x + 1)^{3/2} + C \]

### Final Answer
Thus, the integral of \(\sqrt{x + 1}\) with respect to \(x\) is:
\[ \int \sqrt{x + 1} \, dx = \frac{2}{3} (x + 1)^{3/2} + C \]

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