`int_(0)^(1) x^(3) tan^(-1) x " "dx =`

Options

A. `1/3`
B. `1/6`
C. 3
D. 6

Answer

To solve the integral \( I = \int_{0}^{1} x^3 \tan^{-1}(x) \, dx \), we will use the method of integration by parts.

### Step 1: Identify \( u \) and \( dv \)
We will choose:
- \( u = \tan^{-1}(x) \) (which we will differentiate)
- \( dv = x^3 \, dx \) (which we will integrate)

### Step 2: Differentiate \( u \) and Integrate \( dv \)
Now, we compute \( du \) and \( v \):
- Differentiate \( u \):
\[ du = \frac{1}{1+x^2} \, dx \]
- Integrate \( dv \):
\[ v = \int x^3 \, dx = \frac{x^4}{4} \]

### Step 3: Apply Integration by Parts Formula
Using the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
we substitute our values:
\[ I = \left[ \tan^{-1}(x) \cdot \frac{x^4}{4} \right]_{0}^{1} - \int_{0}^{1} \frac{x^4}{4} \cdot \frac{1}{1+x^2} \, dx \]

### Step 4: Evaluate the Boundary Term
Now we evaluate the boundary term:
\[ \left[ \tan^{-1}(x) \cdot \frac{x^4}{4} \right]_{0}^{1} = \left( \tan^{-1}(1) \cdot \frac{1^4}{4} \right) - \left( \tan^{-1}(0) \cdot \frac{0^4}{4} \right) \]
Calculating this gives:
\[ = \left( \frac{\pi}{4} \cdot \frac{1}{4} \right) - 0 = \frac{\pi}{16} \]

### Step 5: Simplify the Integral
Now we need to simplify the remaining integral:
\[ I = \frac{\pi}{16} - \frac{1}{4} \int_{0}^{1} \frac{x^4}{1+x^2} \, dx \]
We can rewrite \( \frac{x^4}{1+x^2} \) as:
\[ \frac{x^4}{1+x^2} = \frac{x^4}{1+x^2} = x^2 - \frac{x^2}{1+x^2} \]
Thus, we can split the integral:
\[ \int_{0}^{1} \frac{x^4}{1+x^2} \, dx = \int_{0}^{1} x^2 \, dx - \int_{0}^{1} \frac{x^2}{1+x^2} \, dx \]
Calculating these integrals:
1. \( \int_{0}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{1}{3} \)
2. For \( \int_{0}^{1} \frac{x^2}{1+x^2} \, dx \), we can use substitution or recognize it as:
\[ \int_{0}^{1} \frac{x^2}{1+x^2} \, dx = \int_{0}^{1} \left( 1 - \frac{1}{1+x^2} \right) \, dx = 1 - \frac{\pi}{4} \]

### Step 6: Combine Results
Thus, we have:
\[ \int_{0}^{1} \frac{x^4}{1+x^2} \, dx = \frac{1}{3} - \left( 1 - \frac{\pi}{4} \right) = \frac{1}{3} - 1 + \frac{\pi}{4} = \frac{\pi}{4} - \frac{2}{3} \]

### Step 7: Substitute Back into \( I \)
Substituting back into our expression for \( I \):
\[ I = \frac{\pi}{16} - \frac{1}{4} \left( \frac{\pi}{4} - \frac{2}{3} \right) \]
This simplifies to:
\[ I = \frac{\pi}{16} - \frac{\pi}{16} + \frac{1}{6} = \frac{1}{6} \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{\frac{1}{6}} \]