Ex 7.5, 7 - Chapter 7 Class 12 Integrals
Last updated at April 16, 2024 by Teachoo
Ex 7.5
Ex 7.5, 2
Ex 7.5, 3 Important
Ex 7.5, 4
Ex 7.5, 5
Ex 7.5, 6 Important
Ex 7.5, 7 Important You are here
Ex 7.5, 8
Ex 7.5, 9 Important
Ex 7.5, 10
Ex 7.5, 11 Important
Ex 7.5, 12
Ex 7.5, 13 Important
Ex 7.5, 14 Important
Ex 7.5, 15
Ex 7.5, 16 Important
Ex 7.5, 17
Ex 7.5, 18 Important
Ex 7.5, 19
Ex 7.5, 20 Important
Ex 7.5, 21 Important
Ex 7.5, 22 (MCQ)
Ex 7.5, 23 (MCQ) Important
Last updated at April 16, 2024 by Teachoo
Ex 7.5, 7 /( ^2 + 1)( 1) Let I= 1 /( ^2 + 1)( 1) We can write integrand as /( ^2 + 1)( 1) =( + )/(( ^2 + 1) ) + /(( 1) ) /( ^2 + 1)( 1) =(( + )( 1) + ( ^2 + 1))/( ^2 + 1)( 1) By Cancelling denominator =( + )( 1)+ ( ^2+1) Putting x = 1, in (1) =( + )( 1)+ ( ^2+1) 1=( 1+ )(1 1)+ (1^2+1) 1=( + ) 0+ (1+1) 1=0+2 =1/2 Putting x = 0 , in (1) =( + )( 1)+ ( ^2+1) 0=( 0+ )(0 1)+ (0+1) 0 = ( 1)+ = =1/2 Putting x = 1 1=( ( 1)+ )( 1 1)+ ((1)^2+1) 1=( + )( 2)+ (1+1) 1=2 2 +2 1=2 2 1/2+2 1/2 1=2 1+1 1=2 =( 1)/2 Hence we can write /( ^2 + 1)( 1) =(( 1/2 + 1/2))/(( ^2 + 1) ) + (1/2)/(( 1) ) =( 1 . )/(2 ( ^2 + 1) ) + 1/(2 ( ^2 + 1) )+1/2( 1) Integrating . . . I= 1 /( ^2 + 1)( 1) = 1/2 1 /( ^2 + 1) +1/2 1 /( ^2 + 1) + 1/2 1 /( 1) Solving I1= 1/2 1 /(( ^2 + 1) ) Put ^2+1= Different both sides w.r.t. 2 +0= / = /2 Hence we can write 1/2 1 /(( ^2 + 1) ) = 1/2 1 / /2 = 1/(2 2) 1 / = 1/4 log | |+ 1 = 1/4 log | ^2+1|+ 1 Solving I2=1/2 1 1/( ^2 + 1) =1/2 tan^( 1) + 2 Solving I3=1/2 1 1/( 1) =1/2 log | 1|+ 3 Hence I=I1+I2+I3 = 1/4 log | ^2+1|+ 1+1/2 tan^( 1) + 2+1/2 log | 1|+ 3 = 1/4 log | ^2+1|+1/2 tan^( 1) +1/2 log | 1|+ = 1/4 log | ^2+1|+1/2 tan^( 1) +1/2 log | 1|+ = / | | / | ^ + |+ / ^( ) + C