Question

Evaluate : \(\displaystyle\int_{0}^{1} x\tan^{-1}x\,dx\).

Hint:

Use integration by parts with u = tan^{-1}x and dv = x dx, then split x^2/(1+x^2) as 1 - 1/(1+x^2).

Answer 1

Step 1: Choose the method:
The integrand is a product of x and \(\tan^{-1}x\). We use integration by parts: \(\int u\,dv=uv-\int v\,du\).
We take \(u=\tan^{-1}x\) because its derivative is simple. We take \(dv=x\,dx\).

Step 2: Find du and v:
\[ du=\frac{dx}{1+x^2},\qquad v=\frac{x^2}{2} \]

Step 3: Apply the formula:
\[ \int_0^1 x\tan^{-1}x\,dx=\left[\frac{x^2}{2}\tan^{-1}x\right]_0^1-\frac12\int_0^1\frac{x^2}{1+x^2}\,dx \]
The first part is \(\frac12\cdot\frac{\pi}{4}-0=\frac{\pi}{8}\).

Step 4: Simplify the remaining integral:
Write \(\frac{x^2}{1+x^2}=1-\frac{1}{1+x^2}\).
\[ \int_0^1\frac{x^2}{1+x^2}\,dx=\left[x-\tan^{-1}x\right]_0^1=1-\frac{\pi}{4} \]

Step 5: Put the parts together:
\[ \frac{\pi}{8}-\frac12\left(1-\frac{\pi}{4}\right)=\frac{\pi}{8}-\frac12+\frac{\pi}{8}=\frac{\pi}{4}-\frac12 \]

Final Answer:
The value of the integral is given below. \[ \boxed{\dfrac{\pi}{4}-\dfrac{1}{2}} \]