Evaluate `I(b)=int_(0)^(1)(x^(b))dx=int_(0)^(1)(x^(b)-1)/("ln"x)dx,bge0`.

Answer

To evaluate the integral \( I(b) = \int_0^1 x^b \, dx \) and show that it is equal to \( \int_0^1 \frac{x^b - 1}{\ln x} \, dx \) for \( b \geq 0 \), we will follow these steps:

### Step 1: Evaluate \( I(b) \)
We start with the integral:
\[ I(b) = \int_0^1 x^b \, dx \]
To evaluate this integral, we use the formula for the integral of a power function:
\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]
Thus, we have:
\[ I(b) = \left[ \frac{x^{b+1}}{b+1} \right]_0^1 \]
Evaluating this from 0 to 1 gives:
\[ I(b) = \frac{1^{b+1}}{b+1} - \lim_{x \to 0^+} \frac{x^{b+1}}{b+1} \]
Since \( b \geq 0 \), as \( x \to 0 \), \( x^{b+1} \to 0 \). Therefore:
\[ I(b) = \frac{1}{b+1} \]

### Step 2: Differentiate \( I(b) \)
Next, we differentiate \( I(b) \) with respect to \( b \):
\[ \frac{d}{db} I(b) = \frac{d}{db} \left( \frac{1}{b+1} \right) = -\frac{1}{(b+1)^2} \]

### Step 3: Relate to the second integral
Now we consider the integral:
\[ \int_0^1 \frac{x^b - 1}{\ln x} \, dx \]
We can differentiate this integral with respect to \( b \):
\[ \frac{d}{db} \int_0^1 \frac{x^b - 1}{\ln x} \, dx = \int_0^1 \frac{\partial}{\partial b} \left( \frac{x^b - 1}{\ln x} \right) \, dx = \int_0^1 \frac{x^b \ln x}{\ln x} \, dx = \int_0^1 x^b \, dx = I(b) \]

### Step 4: Set up the equation
From the differentiation, we have:
\[ \frac{d}{db} I(b) = \int_0^1 \frac{x^b - 1}{\ln x} \, dx \]
Thus, we can equate the two expressions:
\[ -\frac{1}{(b+1)^2} = \int_0^1 \frac{x^b - 1}{\ln x} \, dx \]

### Step 5: Integrate back to find \( I(b) \)
Integrating both sides with respect to \( b \):
\[ I(b) = -\int \frac{1}{(b+1)^2} \, db = -\left(-\frac{1}{b+1}\right) + C = \frac{1}{b+1} + C \]

### Step 6: Determine the constant \( C \)
To find the constant \( C \), we can evaluate \( I(0) \):
\[ I(0) = \int_0^1 x^0 \, dx = \int_0^1 1 \, dx = 1 \]
From our expression:
\[ I(0) = \frac{1}{0+1} + C = 1 + C \]
Setting this equal to 1 gives:
\[ 1 + C = 1 \implies C = 0 \]

### Final Result
Thus, we have:
\[ I(b) = \frac{1}{b+1} \]
And we also find that:
\[ I(b) = \int_0^1 \frac{x^b - 1}{\ln x} \, dx \]
So the final result is:
\[ I(b) = \ln(b+1) \]