Solve all your doubts with Teachoo Black (new monthly pack available now!)
Are you in school? Do you love Teachoo?
We would love to talk to you! Please fill this form so that we can contact you
Integration by partial fraction - Type 5
Integration by partial fraction - Type 5
Last updated at Aug. 20, 2021 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