Intgerate `intx^2/(xsinx+cosx)^2dx`

Answer

To solve the integral \( I = \int \frac{x^2}{(x \sin x + \cos x)^2} \, dx \), we can follow these steps:

### Step 1: Substitution
Let \( t = x \sin x + \cos x \). Then, we need to find \( dt \):
\[ dt = (x \cos x + \sin x) \, dx \]
From this, we can express \( dx \) in terms of \( dt \):
\[ dx = \frac{dt}{x \cos x + \sin x} \]

### Step 2: Rewrite the Integral
Now we can rewrite the integral \( I \):
\[ I = \int \frac{x^2}{t^2} \cdot \frac{dt}{x \cos x + \sin x} \]
This simplifies to:
\[ I = \int \frac{x^2}{t^2 (x \cos x + \sin x)} \, dt \]

### Step 3: Integration by Parts
We will use integration by parts. Let:
- \( u = x \sec x \)
- \( dv = \frac{dt}{t^2} \)

Then, we have:
\[ du = \left(\sec x + x \sec x \tan x\right) \, dx \]
And integrating \( dv \):
\[ v = -\frac{1}{t} \]

### Step 4: Apply Integration by Parts
Using integration by parts:
\[ I = uv - \int v \, du \]
Substituting our values:
\[ I = -\frac{x \sec x}{t} - \int -\frac{1}{t} \left(\sec x + x \sec x \tan x\right) \, dx \]

### Step 5: Substitute Back
Now substitute back \( t = x \sin x + \cos x \):
\[ I = -\frac{x \sec x}{x \sin x + \cos x} + \int \frac{\sec x + x \sec x \tan x}{x \sin x + \cos x} \, dx \]

### Step 6: Final Integration
The integral \( \int \sec^2 x \, dx \) can be computed, which gives:
\[ \int \sec^2 x \, dx = \tan x + C \]

### Final Result
Combining everything, we get:
\[ I = -\frac{x \sec x}{x \sin x + \cos x} + \tan x + C \]