∫x+2x2+2x+3dx is equal to

Let x+2=Addxx2+2x+3+B ⇒     x+2=A(2x+2)+B⇒        x+2=2Ax+(2A+B)On equating the coefficient of x and constant term on both sides, we get      2A=1⇒A=12and 2A+B=2⇒2×12+B=2⇒     B=2-1=1 ⇒    x+2=122x+2+1∴∫x+2x2+2x+3dx=∫12(2x+2)+1x2+2x+3dx                                    =12∫2x+2x2+2x+3dx+∫dxx2+2x+3Let  l1=∫2x+2x2+2x+3dx and I2=∫dxx2+2x+3Then,∫x+2x2+2x+3dx=12I1+I2          .....(i)Now,   I1=∫2x+2x2+2x+3dxLet x2+2x+3=t⇒(2x+2)dx=dt      I2=∫dtt=∫t−1/2dt=t(−1/2)+11/2+1+C=2t+C1   =2x2+2x+3+C1and I2=∫dxx2+2x+3=∫dxx2+2x+(1)2+3−(1)2   =∫dx(x+1)2+(2)2Let x+1=t⇒dx=dtI2=∫dtt2+(2)2=log⁡t+t2+2+C2                          ∵∫dxx2+a2=log∣x+x2+a2=log⁡x+1+(x+1)2+2+C2=log⁡x+1+x2+2x+3+C2On putting the values of l1 and l2 in Eq. (i), we get∫x+2x2+2x+3dx=122x2+2x+3+log⁡x+1+x2+2x+3∵12C1+C2=C=x2+2x+3+log∣x+1+x2+2x+3+C