Ex 7.5, 9 - Chapter 7 Class 12 Integrals

Transcript

Ex 7.5, 9 3 + 5 3 2 + 1 Let I= 3 + 5 3 2 + 1 We can write integrand as 3 + 5 3 2 + 1 = 3 + 5 1 2 1 = 3 + 5 1 1 + 1 = 3 + 5 + 1 1 2 We can write it as 3 + 5 + 1 1 2 = + 1 + 1 + 1 2 3 + 5 + 1 1 2 = 1 2 + + 1 1 + + 1 + 1 1 1 3 + 5 + 1 1 2 = 1 2 + 2 1 + + 1 + 1 1 2 By Cancelling denominator 3 +5= 1 2 + 2 1 + +1 Put =1 , in (1) 3 1+5= 1 1 2 + 1 2 1 + 1+1 8= 0+ 0+ 2 8=2 =4 Similarly put = 1 , in (1) 3 +5= 1 2 + 2 1 + +1 3 1 +5= 1 1 2 + 1 2 1 + 1+1 3+5= 2 2 + 1 1 + 0 2=4 + 0+ 0 2=4 = 1 2 Putting x = 0 , in (1) 3 0 +5= 0 1 2 + 0 1 + 0+1 5= + 5= 1 2 +4 5= + 9 2 = 9 2 5 = 1 2 Hence we can write 3 + 5 + 1 1 2 = 1 2 + 1 + 1 2 1 + 4 1 2 3 + 5 + 1 1 2 = 1 2 + 1 1 2 1 + 4 1 2 Integrating . . . I= 3 + 5 3 2 + 1 = 1 2 + 1 1 2 1 + 4 1 2 = 1 2 + 1 1 2 1 +4 1 2 Hence I=I1 I2+I3 Now, I1= 1 2 1 + 1 = 1 2 log +1 + 1 Also, I2 = 1 2 1 1 = 1 2 log 1 + 2 And, I3= 4 1 2 =4 1 1 2 =4 1 2 = 4 1 2 + 1 2 + 1 + 3 = 4 1 1 1 + 3 = 4 1 + 3 Therefore I=I1 I2+I3 I= 1 2 log +1 + 1 1 2 log 1 2+ 4 1 + 3 = 1 2 log +1 + 1 2 log 1 4 1 + 1 2+ 3 = 1 2 log +1 log 1 4 1 + = + +