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Mathematics Question on integral

Integrate the function: x2+x+1(x+1)2(x+2)\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}

Answer

Let  x2+x+1(x+1)2(x+2)=A(x+1)+B(x+1)2+C(x+2)...(1)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)

⇒x2+x+1=A(x+1)(x+2)+B(x+2)+C(x2+2x+1)⇒x^{2}+x+1=A(x+1)(x+2)+B(x+2)+C(x^{2}+2x+1)

⇒x2+x+1=A(x2+3x+2)+B(x+2)+C(x2+2x+1)⇒x^{2}+x+1=A(x^{2}+3x+2)+B(x+2)+C(x^{2}+2x+1)

⇒x2+x+1=(A+C)x2+(3A+B+2C)x+(2A+2B+C)⇒x^{2}+x+1=(A+C)x^{2}+(3A+B+2C)x+(2A+2B+C)

Equating the coefficients of x2,x,x2,x, and constant term,we obtain

A+C=1A+C=1

3A+B+2C=13A+B+2C=1

2A+2B+C=12A+2B+C=1

On solving these equations,we obtain

A=−2,B=1,andC=3A=-2,B=1,and C=3

From equation(1),we obtain

x2+x+1(x+1)2(x+2)=−2(x+1)+3(x+2)+1(x+1)2\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}=\frac{-2}{(x+1)}+\frac{3}{(x+2)}+\frac{1}{(x+1)^{2}}

∫x2+x+1(x+1)2(x+2)dx=−2∫1x+1dx+3∫1(x+2)dx+∫1(x+1)2dx\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∣−1(x+1)+C=-2log|x+1|+3log|x+2|-\frac{1}{(x+1)}+C