Let \(I\) =\(∫\)\(xtan^{-1}x\ dx\)
Taking as first function and x as second function and integrating by parts, we obtain
\(I\) = tan-1x∫x dx-∫{(\(\frac {d}{dx}\)tan-1x)∫x dx}dx
\(I\)= tan-1x (\(\frac {x^2}{2}\))-\(∫\frac {1}{1+x^2}.\frac {x^2}{2} dx\)
\(I\)= \(\frac {x^2tan^{-1}x}{2}\) - \(\frac 12\)\(∫\frac {x^2}{1+x^2} dx\)
\(I\)= \(\frac {x^2tan^{-1}x}{2}\) - \(\frac 12\)\(∫(\frac {x^2+1}{1+x^2}-\frac {1}{1+x^2})dx\)
\(I\)= \(\frac {x^2tan^{-1}x}{2}\) - \(\frac 12\)\(∫(1-\frac {1}{1+x^2})dx\)
\(I\)= \(\frac {x^2tan^{-1}x}{2}\) - \(\frac 12\)\((x-tan^{-1}x)+C\)
\(I\)= \(\frac {x^2}{2}tan^{-1}x - \frac x2+\frac 12tan^{-1}x+C\)
Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]
"There is widely spatial variation in different sectors of work participation in India." Evaluate the statement with suitable examples.
Alexia Limited invited applications for issuing 1,00,000 equity shares of ₹ 10 each at premium of ₹ 10 per share.
The amount was payable as follows:
Applications were received for 1,50,000 equity shares and allotment was made to the applicants as follows:
Category A: Applicants for 90,000 shares were allotted 70,000 shares.
Category B: Applicants for 60,000 shares were allotted 30,000 shares.
Excess money received on application was adjusted towards allotment and first and final call.
Shekhar, who had applied for 1200 shares failed to pay the first and final call. Shekhar belonged to category B.
Pass necessary journal entries for the above transactions in the books of Alexia Limited. Open calls in arrears and calls in advance account, wherever necessary.
The number of formulas used to decompose the given improper rational functions is given below. By using the given expressions, we can quickly write the integrand as a sum of proper rational functions.

For examples,
