∫5x−21+2x+3x2dx is equal to
56log1+2x+3x2−1132tan−13x+12+C
53log1+2x+3x2−113tan−13x+12+C
13log1+2x+3x2−1132tan−13x+12+C
None of the above
∫5x−21+2x+3x2dxLet 5x−2=Addx1+2x+3x2+B⇒5x−2=A(2+6x)+B⇒5x−2=6Ax+(2A+B)On equating the coefficient of x and constant on both sides, we get5=6A⇒A=56and 2A+B=−2⇒B=−113∴ l=∫5x−21+2x+3x2dx=∫36(2+6x)−1131+2x+3x2dx=56∫2+6x1+2x+3x2dx−113∫dx1+2x+3x2Let I1=∫2+6x1+2x+3x2 and I2=∫dx1+2x+3x2∴ I=56l1−113I2 ....(i) Now, I1=∫2+6x1+2x+3x2dxLet 1+2x+3x2=t⇒(2+6x)dx=dt⇒ I1=∫dtt=log|t|+C1∴ I1=log1+2x+3x2+C1 ....(ii)Also, I2=∫dx1+2x+3x21+2x+3x2 can be written as 1+3x2+23x. Therefore,1+3x2+23x=1+3x2+23x+19−19 =1+3x+132−13=23+3x+132 =3x+132+29=3x+132+232∴I2=13∫dxx+132+232=13∣12/3tan−1x+132/3+C2∵∫dxa2+x2=1atan−1xa=1332tan−13x+12+C2=12tan−13x+12+C2 ....(iii) On substituting the values of l1 and l2 from Eq. (ii) and Eq. (i),we getI=56log1+2x+3x2−11312tan−13x+12+C∵56C1−113C2=C=56log1+2x+3x2−1132tan−13x+12+C