`int(tan^(-1)x)/(1+x^(2))dx=`

Options

A. `log(tan^(-1)x)+c`
B. `0.5(tan^(-1)x)^(2)+c`
C. `log(1+x^(2))+c`
D. `tan^(-1)(logx)+c`

Answer

To solve the integral \( \int \frac{\tan^{-1} x}{1+x^2} \, dx \), we can use substitution. Here are the 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: Express \( x \) in terms of \( t \)
From the substitution \( t = \tan^{-1} x \), we can express \( x \) as:
\[ x = \tan t \]

### Step 3: Substitute into the integral
Now we substitute \( t \) and \( dx \) into the integral:
\[ \int \frac{\tan^{-1} x}{1+x^2} \, dx = \int \frac{t}{1+\tan^2 t} (1+\tan^2 t) \, dt \]
Since \( 1 + \tan^2 t = \sec^2 t \), the integral simplifies to:
\[ \int t \, dt \]

### Step 4: Integrate
Now we can integrate:
\[ \int t \, dt = \frac{t^2}{2} + C \]

### Step 5: Substitute back
Now, substitute back \( t = \tan^{-1} x \):
\[ \frac{(\tan^{-1} x)^2}{2} + C \]

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