Evaluate :
(i) `int_(0)^(1) (xtan^(-1)x)/((1+x^(2))^(3//2))dx`

Answer

To evaluate the integral

\[ I = \int_{0}^{1} \frac{x \tan^{-1}(x)}{(1+x^2)^{3/2}} \, dx, \]

we will use the substitution \( t = \tan^{-1}(x) \).

### Step-by-Step Solution:

**Step 1: Substitution**
Let \( t = \tan^{-1}(x) \). Then, we have:
\[ x = \tan(t) \quad \text{and} \quad dx = \frac{1}{1+x^2} \, dt. \]
Also, note that \( 1 + x^2 = 1 + \tan^2(t) = \sec^2(t) \).

**Step 2: Change of Limits**
When \( x = 0 \), \( t = \tan^{-1}(0) = 0 \).
When \( x = 1 \), \( t = \tan^{-1}(1) = \frac{\pi}{4} \).

**Step 3: Substitute in the Integral**
Now, substituting these into the integral, we get:
\[ I = \int_{0}^{\frac{\pi}{4}} \frac{\tan(t) \cdot t}{(\sec^2(t))^{3/2}} \cdot \frac{1}{\sec^2(t)} \, dt. \]

**Step 4: Simplify the Integral**
The term \( (\sec^2(t))^{3/2} = \sec^3(t) \), thus:
\[ I = \int_{0}^{\frac{\pi}{4}} \frac{\tan(t) \cdot t}{\sec^3(t)} \cdot \frac{1}{\sec^2(t)} \, dt = \int_{0}^{\frac{\pi}{4}} t \sin(t) \cos(t) \, dt. \]

**Step 5: Use Integration by Parts**
Let \( u = t \) and \( dv = \sin(t) \cos(t) \, dt \). Then, \( du = dt \) and \( v = \frac{1}{2} \sin^2(t) \).

Using integration by parts:
\[ I = \left[ u v \right]_{0}^{\frac{\pi}{4}} - \int_{0}^{\frac{\pi}{4}} v \, du. \]

Evaluating the first term:
\[ \left[ t \cdot \frac{1}{2} \sin^2(t) \right]_{0}^{\frac{\pi}{4}} = \frac{1}{2} \cdot \frac{\pi}{4} \cdot \left( \sin^2\left(\frac{\pi}{4}\right) \right) - 0 = \frac{\pi}{8}. \]

**Step 6: Evaluate the Remaining Integral**
Now we need to evaluate:
\[ \int_{0}^{\frac{\pi}{4}} \frac{1}{2} \sin^2(t) \, dt. \]
Using the identity \( \sin^2(t) = \frac{1 - \cos(2t)}{2} \):
\[ \int_{0}^{\frac{\pi}{4}} \frac{1}{2} \sin^2(t) \, dt = \frac{1}{4} \int_{0}^{\frac{\pi}{4}} (1 - \cos(2t)) \, dt = \frac{1}{4} \left[ t - \frac{1}{2} \sin(2t) \right]_{0}^{\frac{\pi}{4}}. \]

Evaluating this gives:
\[ = \frac{1}{4} \left( \frac{\pi}{4} - 0 - \frac{1}{2} \left( \sin\left(\frac{\pi}{2}\right) - \sin(0) \right) \right) = \frac{1}{4} \left( \frac{\pi}{4} - \frac{1}{2} \right). \]

**Final Calculation: Combine Results**
Combining both parts:
\[ I = \frac{\pi}{8} - \frac{1}{4} \left( \frac{\pi}{4} - \frac{1}{2} \right) = \frac{\pi}{8} - \frac{\pi}{16} + \frac{1}{8} = \frac{2\pi}{16} - \frac{\pi}{16} + \frac{2}{16} = \frac{\pi + 2}{16}. \]

Thus, the final answer is:
\[ I = \frac{\pi + 2}{16}. \]