\(\frac{3x+5}{x^3-x^2-x+1}\) = \(\frac{3x+5}{(x-1)2(x+1)}\)
Let \(\frac{3x+5}{(x-1)^2(x+1)}\) = \(\frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x+1)}\)
3x+5 = A(x-1)(x+1)+B(x+1)+C(x-1)2
3x+5 = A(x2-1)+B(x+1)+C(x2+1-2x) ...(1)
Substituting x = 1 in equation (1), we obtain
B = 4
Equating the coefficients of x2 and x, we obtain
A + C = 0
B − 2C = 3
On solving, we obtain
A = -\(\frac{1}{2}\) and C = \(\frac{1}{2}\)
∴ \(\frac{3x+5}{(x-1)^2(x+1)} = \frac{-1}{2(x-1)}+\frac{4}{(x-1)^2}+\frac{1}{2(x+1)}\)
\(\Rightarrow \int \frac{3x+5}{(x-1)^2(x+1)}dx = -\frac{1}{2}\int\frac{1}{x-1}dx+4\int\frac{1}{(x-1)^2}dx+\frac{1}{2}\int\frac{1}{(x+1)}dx\)
=\(-\frac{1}{2}\log\mid x-1\mid+4\bigg(\frac{-1}{x-1}\bigg)+\frac{1}{2}\log\mid x+1\mid+C\)
=\(\frac{1}{2}\log\mid\frac{x+1}{x-1}\mid-\frac{4}{(x-1)}+C\)
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,
