Evaluate (x−5)x2+xdx=(x2+x)323-Ax+12x+122−122 +Blog⁡x+12+x+122−122+C

Let x−5=λddxx2+x+μ . Then,  x−5=λ2x+1+μ   Comparing coefficients of like powers of x , we get 1=2λ and  λ+μ=−5 or λ=12 and μ=−112, Therefore, ∫(x−5)x2+xdx=∫12(2x+1)−112x2+xdx=12∫(2x+1)x2+xdx−112∫x2+xdx=12∫tdt−112∫x+122−122dx  where t=x2+x=12t3/23/2−11212x+12x+122−122 +12122log⁡x+12+x+122−122+C=(x2+x)323-114x+12x+122−122 -118log⁡x+12+x+122−122+C