\(∫\frac {dx}{x(x^2+1)} \ equals \)
\(log|x|-\frac 12log(x^2+1)+C\)
\(log|x|+\frac 12log(x^2+1)+C\)
\(-log|x|+\frac 12log(x^2+1)+C\)
\(\frac 12log|x|+log(x^2+1)+C\)
Let \(\frac {1}{x(x^2+1)}\) = \(\frac Ax+\frac {Bx+C}{x^2+1}\)
\(1 = A(x^2+1)+(Bx+C)x\)
Equating the coefficients of x2, x, and constant term, we obtain
A + B = 0
C = 0
A = 1
On solving these equations, we obtain
A = 1, B = −1, and C = 0
∴ \(\frac {1}{x(x^2+1)}\) = \(\frac 1x+\frac {-x}{x^2+1}\)
⇒ \(∫\)\(\frac {1}{x(x^2+1)} dx\) = \(∫\)\(\frac 1x-\frac {x}{x^2+1}dx\)
= \(log|x|-\frac 12log|x^2+1|+C\)
Hence, the correct Answer is (A).
Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]
Bittu and Chintu were partners in a firm sharing profit and losses in the ratio of 4 : 3. Their Balance Sheet as at 31st March, 2024 was as follows:
On 1st April, 2024, Diya was admitted in the firm for \( \frac{1}{7} \)th share in the profits on the following terms:
Prepare Revaluation Account and Partners' Capital Accounts.
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,
