\(\frac {1}{x(x^4-1)}\)
Multiplying numerator and denominator by x3 , we obtain
\(\frac {1}{x(x^4-1)}\) = \(\frac {x^3}{x^4(x^4-1)}\)
∴ \(∫\)\(\frac {1}{x(x^4-1)}dx\) = \(∫\)\(\frac {x^3}{x^4(x^4-1)}dx\)
Let x4 = t ⇒ 4x3dx = dt
∴ \(∫\)\(\frac {1}{x(x^4-1)}dx\) = \(\frac 14∫\frac {dt}{t(t-1)}\)
Let \(\frac {1}{t(t-1)}\) = \(\frac At+\frac {B}{(t-1)}\)
\(1 = A(t-1) + Bt \) ...(1)
Substituting t = 0 and 1 in (1), we obtain
\(A = -1\ and \ B = 1\)
⇒ \(\frac {1}{t(t-1)}\) = \(\frac {-1}{t}+\frac {1}{t-1}\)
⇒ \(∫\)\(\frac {1}{x(x^4-1)}dx\) = \(\frac 14\) \(∫\)\(\frac {-1}{t}+\frac {1}{t-1}dt\)
= \(\frac 14[-log|t|+log|t-1|]+C\)
= \(\frac 14log\ |\frac {t-1}{t}|+C\)
= \(\frac 14log|\frac {x^4-1}{x^4}|+C\)
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,
