How do you integrate int (2x-2)/ (x^2 - 4x + 13)dx using partial fractions?

1 Answer
maganbhai P.
Mar 7, 2018

I=ln|x^2-4x+13|+2/3*tan^(_1)((x-2)/3)+C

Explanation:

color(red)((1)int(F^'(x))/(F(x))dx=ln|F(x)|+c)
color(red)((2)int1/(x^2+a^2)dx=1/a*tan^(1)(x/a)+c)
Here,
int(2x-2)/(x^2-4x+13)dx=int(2x-4+2)/((x^2-4x+4+9))dx=int(2(x-2)+2)/((x-2)^2+9)dx
take, x-2=trArrdx=dt
I=int(2t+2)/(t^2+9)dt=int(2t)/(t^2+9)dt+2int1/(t^2+3^2)dt
I=int(d/(dt)(t^2+9))/(t^2+9)dt+2*(1/3)*tan^(-1)(t/3)+c
I=ln|t^2+9|+2/3*tan^(-1)(t/3)+C
I=ln|(x-2)^2+9|+2/3*tan^(-1)((x-2)/3)+C
I=ln|x^2-4x+13|+2/3*tan^(_1)((x-2)/3)+C

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