\(\frac {2x}{(x^2+1)(x^2+3)}\)
\(Let \ x^2 = t ⇒ 2x \ dx = dt\)
∴ \(∫\)\(\frac {2x}{(x^2+1)(x^2+3)}dx\) = \(∫\frac {dt}{(t+1)(t+3)}\) ...(1)
\(Let\) \(\frac {dt}{(t+1)(t+3)}\) = \(\frac {A}{(t+1)}+\frac {B}{(t+3)}\)
\(1 = A(t+3)+B(t+1)\) ...(1)
Substituting t = −3 and t = −1 in equation (1), we obtain
\(A = \frac 12\ and \ B = -\frac 12\)
∴ \(\frac {1}{(t+1)(t+3)}\) = \(\frac {1}{2(t+1)}-\frac {1}{2(t+3)}\)
⇒ \(∫\)\(\frac {2x}{(x^2+1)(x^2+3)}dx\) = \(∫\)\([\frac {1}{2(t+1)}-\frac {1}{2(t+3)}]dt\)
=\(\frac 12log\ |(t+1)|-\frac 12log\ |t+3|+C\)
=\(\frac 12\ log|\frac {t+1}{t+3}|+C\)
=\(\frac 12\ log|\frac {x^2+1}{x^2+3}|+C\)
Determine whether each of the following relations are reflexive, symmetric, and transitive.
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
Find the area of the region bounded by the curve y2=x and the lines x=1,x=4 and the x-axis
Find the area of the region bounded by y2=9x, x=2, x=4 and the x-axis in the first quadrant.
Find the area of the region bounded by x2=4y,y=2,y=4 and the x-axis in the first quadrant.
Find the area of the region bounded by the ellipse \(\frac{x^2}{16}+\frac{y^2}{9}=1\)
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,
