`Ifint_0^1cot^(-1)(1-x+x^2)dx=lambdaint_0^1tan^(-1)x dx ,t h e nlambdai se q u a lto` 1 (b) 2 (c) 3 (d) 4

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem, we need to evaluate the integral

\[ \int_0^1 \cot^{-1}(1 - x + x^2) \, dx \]

and relate it to

\[ \lambda \int_0^1 \tan^{-1}(x) \, dx. \]

### Step 1: Rewrite the cotangent inverse

We know that

\[ \cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right). \]

Thus, we can rewrite the integral as:

\[ \int_0^1 \cot^{-1}(1 - x + x^2) \, dx = \int_0^1 \tan^{-1}\left(\frac{1}{1 - x + x^2}\right) \, dx. \]

### Step 2: Simplify the expression inside the tangent inverse

Now, we can express \(1\) as \(x + (1 - x)\):

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

So we have:

\[ \int_0^1 \tan^{-1}\left(\frac{1}{x^2 + (1 - x)}\right) \, dx. \]

### Step 3: Use the property of definite integrals

Using the property of definite integrals, we know:

\[ \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx. \]

In our case, we can apply this property:

\[ \int_0^1 \tan^{-1}\left(\frac{1}{1 - x + x^2}\right) \, dx = \int_0^1 \tan^{-1}\left(\frac{1}{1 - (1 - x) + (1 - x)^2}\right) \, dx. \]

### Step 4: Combine the integrals

Now, we can combine the two integrals:

\[ \int_0^1 \tan^{-1}\left(\frac{1}{1 - x + x^2}\right) \, dx + \int_0^1 \tan^{-1}\left(\frac{1}{x^2 + (1 - x)}\right) \, dx = \lambda \int_0^1 \tan^{-1}(x) \, dx. \]

### Step 5: Recognize the symmetry

Notice that both integrals on the left-hand side are equal. Therefore, we can write:

\[ 2 \int_0^1 \tan^{-1}\left(\frac{1}{1 - x + x^2}\right) \, dx = \lambda \int_0^1 \tan^{-1}(x) \, dx. \]

### Step 6: Solve for \(\lambda\)

From the above equation, we can equate the coefficients:

\[ \lambda = 2. \]

Thus, the value of \(\lambda\) is:

\[ \boxed{2}. \]