`int ((tan^(-1) x )^(3))/( 1+x^(2)) dx ` is equal to a) `3 ( tan^(-1) x )^(2) +C` b) `(1)/(4) ( tan^(-1)x)^(4) +C` c) `( tan^(-1) x)^(4) + C` d) none of these

Options

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

Answer

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, the derivative of \( t \) with respect to \( x \) is given by:
\[ \frac{dt}{dx} = \frac{1}{1 + x^2} \]
This 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: Back Substitute
Now, we substitute back \( t = \tan^{-1} x \):
\[ \frac{(\tan^{-1} x)^4}{4} + C \]

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

### Conclusion
The correct option is:
b) \( \frac{1}{4} (\tan^{-1} x)^4 + C \)

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