How to solve integration #int (2x^2+2x+1)/(x^3+x^2+x) dx# using partial fractions?

Question

1405 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-02T20:07:43

#I=ln|x|+1/2ln|x^2+x+1|+1/sqrt3arc tan((2x+1)/sqrt3)+c#

Here,

#I=int(2x^2+2x+1)/(x^3+x^2+x)dx=int(2x^2+2x+1)/(x(x^2+x+1))dx#

Partial Fraction :

#(2x^2+2x+1)/(x(x^2+x+1))=A/x+(Bx+C)/(x^2+x+1)#

#=>2x^2+2x+1=A(x^2+x+1)+Bx^2+Cx#

#=>2x^2+2x+1=x^2(A+B)+x(A+C)+A#

Comparing coefficients of #x^2 ,x,and# constant term

#A+B=2to(1),A+C=2to(2) andcolor(blue)( A=1to(3)#

Putting #A=1# , into equn.#(1)and (2)#

#1+B=2 and1+C=2#

#=>color(blue)(B=1 and C=1#

So,

#I=int[1/x+(x+1)/(x^2+x+1)]dx#

#=>I=int[1/x+1/2*(2x+2)/(x^2+x+1)]dx#

#=>I=int[1/x+1/2*(2x+1)/(x^2+x+1)+1/2*1/(x^2+x+1)]dx#

#=lnx+1/2int(d(x^2+x+1))/(x^2+x+1)+1/2int1/(x^2+x+1/4+3/4)dx#

#=ln|x|+1/2ln|x^2+x+1|+1/2int1/((x+1/2)^2+(sqrt3/2)^2)dx#

#=ln|x|+1/2ln|x^2+x+1|+1/2*1/(sqrt3/2)arc tan((x+1/2)/(sqrt3/2))+c#

#=ln|x|+1/2ln|x^2+x+1|+1/sqrt3arc tan((2x+1)/sqrt3)+c#