\(Let \space \frac{x^{2+}x+1}{(x+1)^{2}(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x+1)^{2}}+\frac{C}{(x+2)}...(1)\)
\(⇒x^{2}+x+1=A(x+1)(x+2)+B(x+2)+C(x^{2}+2x+1)\)
\(⇒x^{2}+x+1=A(x^{2}+3x+2)+B(x+2)+C(x^{2}+2x+1)\)
\(⇒x^{2}+x+1=(A+C)x^{2}+(3A+B+2C)x+(2A+2B+C)\)
Equating the coefficients of \(x2,x,\) and constant term,we obtain
\(A+C=1\)
\(3A+B+2C=1\)
\(2A+2B+C=1\)
On solving these equations,we obtain
\(A=-2,B=1,and C=3\)
From equation(1),we obtain
\(\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}=\frac{-2}{(x+1)}+\frac{3}{(x+2)}+\frac{1}{(x+1)^{2}}\)
\(\int \frac{x^{2}+x+1}{(x+1)^{2}(x+2)}dx=-2\int\frac{1}{x+1}dx+3\int\frac{1}{(x+2)}dx+\int\frac{1}{(x+1)^{2}}dx\)
\(=-2log|x+1|+3log|x+2|-\frac{1}{(x+1)}+C\)
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 \(\frac{x^2}{16}+\frac{y^2}{9}=1\)