We know that,
color(red)((1)int(d(f(x)))/f(x) =ln|f(x)|+c(1)∫d(f(x))f(x)=ln|f(x)|+c
color(blue)((2)int1/(x^2+a^2)dx=1/atan^-1(x/a)+c(2)∫1x2+a2dx=1atan−1(xa)+c
Here,
I=intx^2/(x^2+x+4)dxI=∫x2x2+x+4dx
=int((x^2+x+4)-(x+4))/(x^2+x+4)dx=∫(x2+x+4)−(x+4)x2+x+4dx
=int(1-(x+4)/(x^2+x+4))dx=∫(1−x+4x2+x+4)dx
=int1dx-1/2int(2x+8)/(x^2+x+4)dx=∫1dx−12∫2x+8x2+x+4dx
=x-1/2int((2x+1)+7)/(x^2+x+4)dx=x−12∫(2x+1)+7x2+x+4dx
=x-1/2int(2x+1)/(x^2+x+4)dx-1/2int7/(x^2+x+4)dx=x−12∫2x+1x2+x+4dx−12∫7x2+x+4dx
=x-1/2int(color(red)(d(x^2+x+4))/color(red)((x^2+x+4)))dx-
7/2int1/(x^2+x+1/4+15/4)dx=x−12∫(d(x2+x+4)(x2+x+4))dx−72∫1x2+x+14+154dx
=x-1/2color(red)(ln|x^2+x+4|)-7/2color(blue)(int 1/((x+1/2)^2+
(sqrt15/2)^2)dx=x−12ln∣∣x2+x+4∣∣−72∫1(x+12)2+(√152)2dx
=x-1/2ln|x^2+x+4|-7/2*color(blue)(1/(sqrt15/2)
tan^-1((x+1/2)/(sqrt15/2)))+c=x−12ln∣∣x2+x+4∣∣−72⋅1√152tan−1⎛⎜⎝x+12√152⎞⎟⎠+c
=x-1/2ln|x^2+x+4|-7/sqrt15 tan^-1((2x+1)/sqrt15)+c=x−12ln∣∣x2+x+4∣∣−7√15tan−1(2x+1√15)+c
