Question

Calculate the value of ∫1 (tan-1x/1+x2) dx

• A. -2π
• B. 2-π²/16
• C. π²/32
• D. 2/16

Answer 1

The Correct Option is C

We need to evaluate the integral \( \int_{0}^{1} \frac{\tan^{-1} x}{1 + x^2} \, dx \).

1. Analyze the Integrand:
The integrand is \( \frac{\tan^{-1} x}{1 + x^2} \). Notice that the denominator \( 1 + x^2 \) resembles the derivative of \( \tan^{-1} x \). Let’s compute the derivative of \( \tan^{-1} x \):

2. Derivative of \( \tan^{-1} x \):
If \( y = \tan^{-1} x \), then \( x = \tan y \), and differentiating with respect to \( x \):

\( 1 = \sec^2 y \cdot \frac{dy}{dx} \)
Since \( \sec^2 y = 1 + \tan^2 y = 1 + x^2 \), we have:

\( \frac{dy}{dx} = \frac{1}{1 + x^2} \)
Thus, the derivative of \( \tan^{-1} x \) is \( \frac{1}{1 + x^2} \), so:

\( \frac{d}{dx} (\tan^{-1} x) = \frac{1}{1 + x^2} \)

3. Substitution:
Notice that the denominator of the integrand is \( 1 + x^2 \), which matches the derivative. Let’s try a substitution. Set:

\( u = \tan^{-1} x \)
Then, \( x = \tan u \), and differentiating:

\( \frac{d}{dx} (\tan^{-1} x) = \frac{1}{1 + x^2} \)
So:

\( du = \frac{1}{1 + x^2} \, dx \)
The integrand becomes:

\( \frac{\tan^{-1} x}{1 + x^2} \, dx = \tan^{-1} x \cdot \frac{1}{1 + x^2} \, dx = u \, du \)

4. Adjust the Limits:
The limits of integration are from \( x = 0 \) to \( x = 1 \):
- When \( x = 0 \), \( u = \tan^{-1} 0 = 0 \).
- When \( x = 1 \), \( u = \tan^{-1} 1 = \frac{\pi}{4} \).
The integral transforms to:

\( \int_{x=0}^{1} \frac{\tan^{-1} x}{1 + x^2} \, dx = \int_{u=0}^{\frac{\pi}{4}} u \, du \)

5. Evaluate the Integral:
Now integrate \( u \) with respect to \( u \):

\( \int_{0}^{\frac{\pi}{4}} u \, du = \left[ \frac{u^2}{2} \right]_{0}^{\frac{\pi}{4}} \)
Substitute the limits:

\( \frac{1}{2} \left( \left( \frac{\pi}{4} \right)^2 - 0^2 \right) = \frac{1}{2} \cdot \frac{\pi^2}{16} = \frac{\pi^2}{32} \)

6. Verify:
The antiderivative of \( u \) is \( \frac{u^2}{2} \), and substituting back \( u = \tan^{-1} x \), the antiderivative of the original integrand is \( \frac{1}{2} (\tan^{-1} x)^2 \), which can be differentiated to confirm:

\( \frac{d}{dx} \left( \frac{1}{2} (\tan^{-1} x)^2 \right) = \frac{1}{2} \cdot 2 (\tan^{-1} x) \cdot \frac{1}{1 + x^2} = \frac{\tan^{-1} x}{1 + x^2} \), which matches the integrand.

Final Answer:
The value of the integral is: \( \int_{0}^{1} \frac{\tan^{-1} x}{1 + x^2} \, dx = \frac{\pi^2}{32} \)