∫5x−21+2x+3x2dxlet,u=5x−2∴∫5x−21+2x+3x2dx=∫5u3u2+22u+57du=5∫u3(u+113)2+503duagain,letv=u+113∴5∫u3(u+113)2+503du=5∫3v−119v2+50dvsubstituting,v=5√23w∴∫3v−119v2+50dv=5.∫5.√2w−1115√2(w2+1)dw=515√2(∫5√2w2+1dw−∫11w2+1dw)=515√2[5√2log(35√2((5x−2+113))2)−11arctan35√2((5x−2+113))]+C=16[5log((3x+1)22+1)−11√2arctan(3x+1√2)]+C