Question

Evaluate: \(\displaystyle\int_0^1 \frac{\tan^{-1}x}{1+x^2}\, dx\)

Hint:

Substitute t=tan⁻¹x, since dt=dx/(1+x²), turning it into ∫t dt.

Answer 1

Step 1: Understanding the Concept:
Notice that \(\dfrac{d}{dx}\tan^{-1}x = \dfrac{1}{1+x^2}\). This means the integrand is of the form \(\tan^{-1}x \cdot \dfrac{d}{dx}(\tan^{-1}x)\), which is perfectly set up for the substitution \(t=\tan^{-1}x\).

Step 2: Substituting:
Let \(t=\tan^{-1}x\), so \(dt = \dfrac{dx}{1+x^2}\). The integral becomes:
\[ \int t\,dt \]

Step 3: Changing the limits:
When \(x=0\), \(t=\tan^{-1}0=0\). When \(x=1\), \(t=\tan^{-1}1=\dfrac{\pi}{4}\).

Step 4: Integrating and applying limits:
\[ \int_0^{\pi/4} t\,dt = \left[\frac{t^2}{2}\right]_0^{\pi/4} = \frac{(\pi/4)^2}{2} - 0 = \frac{\pi^2}{32} \]

Final Answer:
The value of the integral is \(\pi^2/32\). \[ \boxed{\dfrac{\pi^2}{32}} \]