If ∫dx(x+1)2x2+3x+4=Klog⁡f(x)+C, then

Putting x+1=1/y, the given integral can be written as −∫dy3y2−y+2=−13∫dy(y−1/6)2+(23/6)2=−13log⁡(y−1/6)+(y−1/6)2+(23/6)2+C=−13log⁡5−x6(x+1)+5−x6(x+1)2+2362+C.