(i) `int (x cos^-1 x)/sqrt(1-x)^2 dx` (ii) `int (x tan^-1x)/(1+x^2)^(3//2) dx`

Answer

Let's solve the two integrals step by step.

### Part (i):
Evaluate the integral
\[ I = \int \frac{x \cos^{-1}(x)}{\sqrt{1 - x^2}} \, dx. \]

**Step 1: Substitution**
Let \( t = \cos^{-1}(x) \). Then, we have:
\[ x = \cos(t) \quad \text{and} \quad dx = -\sin(t) \, dt. \]
Also, note that:
\[ \sqrt{1 - x^2} = \sqrt{1 - \cos^2(t)} = \sin(t). \]
Substituting these into the integral gives:
\[ I = \int \frac{\cos(t) t}{\sin(t)} (-\sin(t)) \, dt = -\int t \cos(t) \, dt. \]

**Step 2: Integration by Parts**
Using integration by parts, where we let:
- \( u = t \) and \( dv = \cos(t) \, dt \),
- then \( du = dt \) and \( v = \sin(t) \).

Applying the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du, \]
we get:
\[ I = -\left( t \sin(t) - \int \sin(t) \, dt \right). \]
Calculating the integral of \( \sin(t) \):
\[ \int \sin(t) \, dt = -\cos(t). \]
Thus,
\[ I = -\left( t \sin(t) + \cos(t) \right) + C. \]

**Step 3: Back Substitution**
Now, substituting back \( t = \cos^{-1}(x) \):
\[ I = -\left( \cos^{-1}(x) \sin(\cos^{-1}(x)) + \cos(\cos^{-1}(x)) \right) + C. \]
Since \( \sin(\cos^{-1}(x)) = \sqrt{1 - x^2} \) and \( \cos(\cos^{-1}(x)) = x \):
\[ I = -\left( \cos^{-1}(x) \sqrt{1 - x^2} + x \right) + C. \]

### Final Result for Part (i):
\[ I = -\cos^{-1}(x) \sqrt{1 - x^2} - x + C. \]

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### Part (ii):
Evaluate the integral
\[ J = \int \frac{x \tan^{-1}(x)}{(1 + x^2)^{3/2}} \, dx. \]

**Step 1: Substitution**
Let \( u = \tan^{-1}(x) \). Then, we have:
\[ x = \tan(u) \quad \text{and} \quad dx = \sec^2(u) \, du. \]
Also, note that:
\[ 1 + x^2 = 1 + \tan^2(u) = \sec^2(u) \quad \text{thus} \quad (1 + x^2)^{3/2} = \sec^3(u). \]
Substituting these into the integral gives:
\[ J = \int \frac{\tan(u) u}{\sec^3(u)} \sec^2(u) \, du = \int u \sin(u) \, du. \]

**Step 2: Integration by Parts**
Using integration by parts again:
- Let \( u = u \) and \( dv = \sin(u) \, du \),
- then \( du = du \) and \( v = -\cos(u) \).

Applying the integration by parts formula:
\[ J = -u \cos(u) + \int \cos(u) \, du. \]
Calculating the integral of \( \cos(u) \):
\[ \int \cos(u) \, du = \sin(u). \]
Thus,
\[ J = -u \cos(u) + \sin(u) + C. \]

**Step 3: Back Substitution**
Now, substituting back \( u = \tan^{-1}(x) \):
\[ J = -\tan^{-1}(x) \cos(\tan^{-1}(x)) + \sin(\tan^{-1}(x)) + C. \]
Since \( \cos(\tan^{-1}(x)) = \frac{1}{\sqrt{1 + x^2}} \) and \( \sin(\tan^{-1}(x)) = \frac{x}{\sqrt{1 + x^2}} \):
\[ J = -\tan^{-1}(x) \cdot \frac{1}{\sqrt{1 + x^2}} + \frac{x}{\sqrt{1 + x^2}} + C. \]

### Final Result for Part (ii):
\[ J = -\frac{\tan^{-1}(x)}{\sqrt{1 + x^2}} + \frac{x}{\sqrt{1 + x^2}} + C. \]

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