Miscellaneous

Chapter 7 Class 12 Integrals
Serial order wise

### Transcript

Misc 14 Integrate the function 1/((π₯^2 + 1)(π₯^2 + 4) ) Integrating 1/(π₯^2 + 1)(π₯^2 + 4) Let π₯^2=π¦ 1/(π₯^2 + 1)(π₯^2 + 4) =1/(π¦ + 4)(π¦ + 1) Hence we can write 1/(π¦ + 4)(π¦ + 1) =π΄/((π¦ + 4) )+π΅/((π¦ + 1) ) 1/(π¦ + 4)(π¦ + 1) =(π΄(π¦ + 1)+π΅(π¦ + 4))/(π¦ + 4)(π¦ + 1) Canceling denominators 1 = A(π¦+1)+B(π¦+4) Putting π=βπ 1= A (β1+1)+B(β1+4) 1= A Γ0+ B (3) 1=3B B =1/3 Putting π=βπ 1= A(β4+1)+ B(β4+4) 1= A(β3)+ B(0) 1=β3A A =(β1)/3 Therefore we can write 1/(π¦ + 4)(π¦ + 1) =π΄/((π¦ + 4) )+π΅/((π¦ + 1) ) 1/(π¦ + 4)(π¦ + 1) =(((β1)/( 3)))/((π¦ + 4) )+((1/3))/((π¦ + 1) ) Putting back π¦=π₯^2 1/(π₯^2 + 4)(π₯^2 + 1) =(((β1)/( 3)))/((π₯^2 + 4) )+((( 1)/( 3)))/((π₯^2+ 1) ) = (β1)/3(π₯^2 + 4) +1/3(π₯^(2 )+ 1) Integrating w.r.t. π₯ β«1ββ(1/(π₯^(2 )+ 4)(π₯^2 + 1) ππ₯) =β«1β((β1)/3(π₯^2 + 4) +1/3(π₯^2 + 1) )ππ₯ =β«1βγ(β1)/3(π₯^2 + 4) ππ₯+γ β«1βγ(β1)/3(π₯^2 + 1) ππ₯γ =(β1)/( 3) β«1βγ1/(π₯^2 + 2^2 ) ππ₯+1/3 β«1βγ1/(π₯^2 + 1^2 ) ππ₯γγ =(β1)/3 Γ 1/2 tan^(β1)β‘γπ₯/2+1/3 Γ 1/1 tan^(β1)β‘γπ₯/1+πΆγ γ =(β1)/( 6) tan^(β1)β‘γπ₯/2+1/3 tan^(β1)β‘γπ₯+πγ γ We know that β«1βγ1/(π₯^2 + π^2 ) ππ₯=1/π tan^(β1)β‘γπ₯/π+πγ γ =π/π γπ­ππ§γ^(βπ)β‘γπβπ/π γπ­ππ§γ^(βπ)β‘γπ/π+πγ γ