`int((1+x)sinx)/((x^(2)+2x)cos^(2)x-(1+x)sin 2x)dx`

Options

A. `(1)/(2)log_(e)|(sinx-(x+1)cosx-1)/(sinx-(x+1)cosx+1)|+C`
B. `(1)/(2)tan^(-1){sinx-(x+1)cosx}+C`
C. `(1)/(2)sin^(-1){sinx-(x+1)cosx}+C`
D. `(1)/(2)sin^(-1)(cosx+sinx)+C`

Explanation

To solve the integral

\[ I = \int \frac{(1+x) \sin x}{(x^2 + 2x) \cos^2 x - (1+x) \sin 2x} \, dx, \]

we will follow these steps:

### Step 1: Simplify the Denominator

The denominator can be rewritten as:

\[ (x^2 + 2x) \cos^2 x - (1+x) \sin 2x. \]

We know that \(\sin 2x = 2 \sin x \cos x\), so we can substitute that in:

\[ = (x^2 + 2x) \cos^2 x - (1+x)(2 \sin x \cos x). \]

### Step 2: Factor the Denominator

Notice that \(x^2 + 2x = (x+1)^2 - 1\). Thus, we can rewrite the denominator as:

\[ ((x+1)^2 - 1) \cos^2 x - (1+x)(2 \sin x \cos x). \]

Now, let's rewrite the denominator:

\[ = (x+1)^2 \cos^2 x - \cos^2 x - 2(1+x) \sin x \cos x. \]

### Step 3: Rewrite the Denominator

Now we can express the denominator in a more manageable form. We will add and subtract \(\cos^2 x\):

\[ = (x+1)^2 \cos^2 x - 2(1+x) \sin x \cos x - 1. \]

### Step 4: Recognize the Form

The denominator can be recognized as a difference of squares:

\[ = ((x+1) \cos x - \sin x)^2 - 1. \]

### Step 5: Substitute

Let

\[ t = (x+1) \cos x - \sin x. \]

Then, the differential \(dt\) can be computed as follows:

\[ dt = [\cos x - (x+1) \sin x] \, dx. \]

### Step 6: Rewrite the Integral

Now we can rewrite the integral \(I\):

\[ I = \int \frac{(1+x) \sin x}{t^2 - 1} \, dx. \]

We can express this as:

\[ I = \int \frac{(1+x) \sin x}{(t-1)(t+1)} \, dx. \]

### Step 7: Use Partial Fraction Decomposition

We can use partial fractions to split the integral:

\[ \frac{(1+x) \sin x}{(t-1)(t+1)} = \frac{A}{t-1} + \frac{B}{t+1}. \]

### Step 8: Integrate

Integrating each term separately gives us:

\[ I = \frac{1}{2} \log \left| \frac{t-1}{t+1} \right| + C. \]

### Step 9: Substitute Back

Finally, substitute back \(t = (x+1) \cos x - \sin x\):

\[ I = \frac{1}{2} \log \left| \frac{(x+1) \cos x - \sin x - 1}{(x+1) \cos x - \sin x + 1} \right| + C. \]

### Final Answer

Thus, the integral evaluates to:

\[ I = \frac{1}{2} \log \left| \frac{(x+1) \cos x - \sin x - 1}{(x+1) \cos x - \sin x + 1} \right| + C. \]