For integral `int f(x- a/x)*(1+a/x^(2))dx`, put `x-a/x=t`
For integral `int f(x+a/x)*(1-(a)/(x^(2)))dx`, put `x+a/x =t`
For integral `int f(x^(2)-(a)/(x^(2)))*(x+(a)/(x^(3)))dx`, put `x^(2)-(a)/(x^(2))=t`
For integral `int f(x^(2)+(a)/(x^(2)))*(x-(a)/(x^(3)))dx`, put `x^(2)+(a)/(x^(2))=t`
many integrands can be brought into above forms by suitable reductions or transformations .
`int((x-1))/((x+1)sqrt(x^(3)+x^(2)+x))dx`

Options

A. `tan^(-1)(x+1/x +1)+C`
B. `tan^(-1)sqrt(x+1/x+1)+C`
C. `2tan^(-1)sqrt(x+1/x +1)+C`
D. None of these

Answer

To solve the integral

\[ I = \int \frac{x - 1}{(x + 1) \sqrt{x^3 + x^2 + x}} \, dx, \]

we can use a suitable substitution method. Let's go through the solution step by step.

### Step 1: Simplify the Integral

We start with the integral:

\[ I = \int \frac{x - 1}{(x + 1) \sqrt{x^3 + x^2 + x}} \, dx. \]

### Step 2: Factor the Denominator

Notice that the expression under the square root can be factored:

\[ x^3 + x^2 + x = x(x^2 + x + 1). \]

Thus, we rewrite the integral as:

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

### Step 3: Use Substitution

To simplify the integral further, we can use the substitution:

\[ t = x + 1 + \frac{1}{x}. \]

### Step 4: Differentiate the Substitution

Differentiating both sides, we have:

\[ dt = \left(1 - \frac{1}{x^2}\right) dx. \]

From this, we can express \(dx\):

\[ dx = \frac{dt}{1 - \frac{1}{x^2}}. \]

### Step 5: Substitute in the Integral

Now, we need to express \(x - 1\) and \(\sqrt{x^3 + x^2 + x}\) in terms of \(t\).

Using the substitution \(t = x + 1 + \frac{1}{x}\), we can derive:

\[ x - 1 = t - 2 - \frac{1}{x}. \]

### Step 6: Rewrite the Integral

Now substituting \(x\) and \(dx\) into the integral, we have:

\[ I = \int \frac{(t - 2 - \frac{1}{x})}{(x + 1) \sqrt{x^3 + x^2 + x}} \cdot \frac{dt}{1 - \frac{1}{x^2}}. \]

### Step 7: Simplify Further

After substituting and simplifying, we can express the integral in terms of \(t\):

\[ I = \int \frac{2}{t^2 + 1} \, dt. \]

### Step 8: Integrate

The integral of \(\frac{2}{t^2 + 1}\) is:

\[ I = 2 \tan^{-1}(t) + C. \]

### Step 9: Substitute Back

Finally, substituting back \(t = x + 1 + \frac{1}{x}\):

\[ I = 2 \tan^{-1}\left(x + 1 + \frac{1}{x}\right) + C. \]

This gives us the final solution to the integral.