intln(x^2+x+1)dx∫ln(x2+x+1)dx
apply Integration by Parts
u=ln(x^2+x+1)u=ln(x2+x+1)
du=(2x+1)/(x^2+x+1)dxdu=2x+1x2+x+1dx
dv=dxdv=dx
v=xv=x
intln(x^2+x+1)dx=xln(x^2+x+1)-int(x(2x+1))/(x^2+x+1)dx∫ln(x2+x+1)dx=xln(x2+x+1)−∫x(2x+1)x2+x+1dx
rarr→(1)(1)
Let I=int(x(2x+1))/(x^2+x+1)dx=I=∫x(2x+1)x2+x+1dx=
simplify
int(2x^2+x)/(x^2+x+1)dx∫2x2+xx2+x+1dx=int(x^2+x+1)/(x^2+x+1)dx+int(x^2-1)/(x^2+x+1)dx∫x2+x+1x2+x+1dx+∫x2−1x2+x+1dx
=x+int(x^2-1)/(x^2+x+1)dx=x+∫x2−1x2+x+1dx
let r=int(x^2-1)/(x^2+x+1)dxr=∫x2−1x2+x+1dx
completing the square of the denominator :
r=int(x^2-1)/((x+1/2)^2+3/4)dxr=∫x2−1(x+12)2+34dx
apply trigonometric substitution
x+1/2=sqrt(3)/2tanthetax+12=√32tanθ
dx=sqrt(3)/2sec^2theta*d(theta)dx=√32sec2θ⋅d(θ)
x^2=3/4tan^2theta-sqrt(3)/2tantheta+1/4x2=34tan2θ−√32tanθ+14
substitute
int(x^2-1)/((x+1/2)^2+3/4)dx=int((3/4tan^2theta-sqrt(3)/2tantheta+1/4)-1)/(3/4tan^2theta+3/4)sqrt(3)/2sec^2thetad(theta)∫x2−1(x+12)2+34dx=∫(34tan2θ−√32tanθ+14)−134tan2θ+34√32sec2θd(θ)
simplify where tan^2theta+1=sec^2thetatan2θ+1=sec2θ
4/3*sqrt(3)/2int(3/4tan^2theta-sqrt(3)/2tantheta-3/4)d(theta)43⋅√32∫(34tan2θ−√32tanθ−34)d(θ)
simplify again using the relation tan^2theta+1=sec^2thetatan2θ+1=sec2θ
2/sqrt3int(3/4sec^2theta-sqrt3/2tantheta-3/2)d(theta)2√3∫(34sec2θ−√32tanθ−32)d(θ)=sqrt3/2tantheta+lncostheta-sqrt3theta+C√32tanθ+lncosθ−√3θ+C
reverse the trigonometric substitution
theta=tan^-1(2/sqrt(3)(x+1/2))θ=tan−1(2√3(x+12))
so,
r=int(x^2-1)/(x^2+x+1)dxr=∫x2−1x2+x+1dx=(x+1/2)+lncostan^-1(2/sqrt(3)(x+1/2))-sqrt3tan^-1(2/sqrt(3)(x+1/2))+C(x+12)+lncostan−1(2√3(x+12))−√3tan−1(2√3(x+12))+C
substituting in II:
I=x+(x+1/2)+lncostan^-1(2/sqrt(3)(x+1/2))-sqrt3tan^-1(2/sqrt(3)(x+1/2))+CI=x+(x+12)+lncostan−1(2√3(x+12))−√3tan−1(2√3(x+12))+C
and substituting with the value of I in (1)
your final integration result will be:
intln(x^2+x+1)dx∫ln(x2+x+1)dx=xln(x^2+x+1)-x-(x+1/2)-lncostan^-1(2/sqrt(3)(x+1/2))+sqrt3tan^-1(2/sqrt(3)(x+1/2))+Cxln(x2+x+1)−x−(x+12)−lncostan−1(2√3(x+12))+√3tan−1(2√3(x+12))+C
