`int((tan^(-1)x)^(3))/(1+x^(2))dx` is equal to

Options

A. `3(tan^(-1)x)^(2)+C`
B. `((tan^(-1)x)^(4))/(4)+C`
C. `(tan^(-1)x)^(4)+C`
D. none of these

Explanation

To solve the integral \( \int \frac{(\tan^{-1} x)^3}{1 + x^2} \, dx \), we can follow these steps:

### Step 1: Substitution
Let \( t = \tan^{-1} x \). Then, we differentiate both sides to find \( dx \):
\[ \frac{dt}{dx} = \frac{1}{1 + x^2} \implies dx = (1 + x^2) \, dt \]

### Step 2: Rewrite the Integral
Substituting \( t \) into the integral, we have:
\[ \int \frac{(\tan^{-1} x)^3}{1 + x^2} \, dx = \int t^3 \, dt \]

### Step 3: Integrate
Now we can integrate \( t^3 \):
\[ \int t^3 \, dt = \frac{t^4}{4} + C \]

### Step 4: Substitute Back
Now we substitute back \( t = \tan^{-1} x \):
\[ \frac{t^4}{4} + C = \frac{(\tan^{-1} x)^4}{4} + C \]

### Final Answer
Thus, the integral is:
\[ \int \frac{(\tan^{-1} x)^3}{1 + x^2} \, dx = \frac{(\tan^{-1} x)^4}{4} + C \]

---