Here,
I=int6/((x-5)(x^2-2x+3))dxI=∫6(x−5)(x2−2x+3)dx
Partial Fractions :
color(blue)(6/((x-5)(x^2-2x+3))=A/(x-5)+(Bx+C)/(x^2-2x+3)6(x−5)(x2−2x+3)=Ax−5+Bx+Cx2−2x+3
=>6=A(x^2-2x+3)+Bx(x-5)+C(x-5)⇒6=A(x2−2x+3)+Bx(x−5)+C(x−5)
=>6=x^2(A+B)+x(-2A-5B+C)+3A-5C⇒6=x2(A+B)+x(−2A−5B+C)+3A−5C
Comparing Coefficient of x^2, x and constant term,
A+B=0=>A=-Bto(1)A+B=0⇒A=−B→(1)
-2A-5B+C=0to(2) −2A−5B+C=0→(2)
3A-5C=6to(3)3A−5C=6→(3)
Subst. A=-BA=−B into (2)(2)
:.2B-5B+C=0=>-3B+C=0=>C=3B
From (3) we get,
3(-B)-5(3B)=6=>-18B=6=>B=-1/3
From, (1) A=-(-1/3)=1/3 and C=3(-1/3)=-1
i.e. color(blue)( A=1/3,B=-1/3,C=-1
So
I=int(1/3)/(x-5)dx+int((-1/3x)-1)/(x^2-2x+3)dx
=1/3int1/(x-5)dx-1/3int(x+3)/(x^2-2x+3)dx
=1/3ln|x-5|-1/3*1/2int(2x+6)/(x^2-2x+3)dx
=1/3ln|x-5|-1/6int(2x-2+8)/(x^2-2x+3)dx
=1/3ln|x-5|-1/6int(2x-2)/(x^2-2x+3)dx-1/6int8/(x^2-2x+3)dx
=1/3ln|x-5|-1/6int(d/(dx)(x^2-2x+3))/(x^2-2x+3)dx
color(white)(.........................).-8/6int1/((x^2-2x+1+2))dx
=1/3ln|x-5|-1/6ln|x^2-2x+3|
color(white)(..............................)-4/3color(red)(int1/((x-1)^2+(sqrt2)^2)dx
=1/3ln|x-5|-1/6ln|x^2-2x+3|-4/3*color(red)(1/sqrt2 tan^-1((x-1)/sqrt2)+ c
I=1/3ln|x-5|-1/6ln|x^2-2x+3|-4/(3sqrt2)tan^-1((x-1)/sqrt2)+c
Note :
color(red)(int1/(x^2+a^2)dx=1/atan^-1(x/a)+c
