Integrate tan inverse x

Steps to Solve

1. Identify the Integral to Solve

We want to find the integral of $ \tan^{-1}(x) $, so we set up the integral as follows:

$$ I = \int \tan^{-1}(x) , dx $$

2. Use Integration by Parts

We apply the integration by parts formula, which states:

$$ \int u , dv = uv - \int v , du $$

We choose:

• $ u = \tan^{-1}(x) $ (thus, $ du = \frac{1}{1+x^2} , dx $)
• $ dv = dx $ (thus, $ v = x $)

3. Apply the Parts Formula

Following the integration by parts formula:

$$ I = x \tan^{-1}(x) - \int x \cdot \frac{1}{1+x^2} , dx $$

4. Simplify the Remaining Integral

Now we simplify the remaining integral:

The integral can be rewritten as:

$$ \int \frac{x}{1+x^2} , dx $$

5. Use Substitution for the Simplified Integral

For the integral $ \int \frac{x}{1+x^2} , dx $, we can use the substitution $ w = 1 + x^2 $, then $ dw = 2x , dx $ or $ dx = \frac{dw}{2x} $:

Thus:

$$ \int \frac{x}{1+x^2} , dx = \frac{1}{2} \int \frac{1}{w} , dw $$

6. Integrate the Simple Expression

The integral of $ \frac{1}{w} $ is:

$$ \frac{1}{2} \ln |w| + C = \frac{1}{2} \ln |1+x^2| + C $$

7. Combine All Parts

Now we combine everything back together:

$$ I = x \tan^{-1}(x) - \left( \frac{1}{2} \ln |1+x^2| + C \right) $$

This can be simplified to:

$$ I = x \tan^{-1}(x) - \frac{1}{2} \ln(1+x^2) + C $$