Examples

Chapter 7 Class 12 Integrals
Serial order wise

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Example 10 Find the following integrals: (i) β«1β(π₯ + 2)/(2π₯^2 + 6π₯ + 5 ) ππ₯ We can write numerator as π₯+2= A π/ππ₯ (2π₯^2+6π₯+5) + B π₯+2= A [4π₯+6]+ B π₯+2=4π΄π₯+6A+B Finding A & B Comparing coefficient of π₯ π₯=4π΄π₯ 1 =4A A=1/4 Comparing constant term 2=6A+B 2=6(1/4)+B 2=3/2+B B=2β3/2=1/2 Now, we know that π₯+2= A [4π₯+6]+ B π₯+2=1/4 [4π₯+6]+1/2 Now, our equation is β«1βγ(π₯ + 2)/(2π₯^2 + 6π₯ + 5).ππ₯=β«1βγ(1/4 [4π₯ + 6] + 1/2)/(2π₯^2 + 6π₯ + 5).ππ₯γγ =β«1βγ(1/4 [4π₯ + 6])/(2π₯^2 + 6π₯ + 5)+β«1βγ(1/2)/(2π₯^2+6π₯+5).ππ₯γγ =1/4 β«1βγ(4π₯ + 6)/(2π₯^2 + 6π₯ + 5) ππ₯+1/2 β«1βγ1/(2π₯^2 + 6π₯ + 5).ππ₯γγ Solving I1 I1 =1/4 β«1βγ(4π₯ + 6)/(2π₯^2 + 6π₯ + 5) ππ₯γ Let t = 2π₯^2 + 6π₯ + 5 Differentiating both sides w.r.t.π₯ 4π₯ +6=ππ‘/ππ₯ ππ₯=ππ‘/(4π₯ + 6) Now, I1 =1/4 β«1βγ(4π₯ + 6)/(2π₯^2 + 6π₯ + 5).ππ₯γ Putting the values of (2π₯^2+6π₯+5) and ππ₯, we get I1 =1/4 β«1βγ(4π₯ + 6)/π‘.ππ‘/(4π₯ + 6) γ I1 =1/4 β«1βγ1/π‘.ππ‘ γ I1 =1/4 πππ|π‘|+πΆ1 I1 =1/4 πππ|2π₯^2+6π₯+5|+πΆ1 Solving I2 I2 =1/2 β«1βγ1/(2π₯^2 + 6π₯ + 5).ππ₯ γ I2 =1/2 β«1βγ1/2[π₯^2 + 6π₯/2 + 5/2 ] .ππ₯ γ I2 =1/4 β«1βγ1/(π₯^2 +3π₯ + 5/2).ππ₯ γ (Using β«1βγ1/π₯.ππ₯=πππ|π₯|+πΆ1γ) (Using π‘=2π₯^2+6π₯+5) I2 =1/4 β«1βγ1/(π₯^2 + 2(π₯)(3/2) + 5/2).ππ₯ γ Adding & subtracting (3/2)^2 in denominator I2 =1/4 β«1βγ1/(π₯^2 + 2(π₯) γ(3/2) +(3/2)^2β (3/2)γ^2 + 5/2).ππ₯ γ I2 =1/4 β«1βγ1/((π₯ + 3/2)^2 β (3/2)^2 + 5/2) ππ₯ γ I2 =1/4 β«1βγ1/((π₯ + 3/2)^2 β 9/4 + 5/2).ππ₯ γ I2 =1/4 β«1βγ1/((π₯ + 3/2)^2+ (β9 + 10)/4 ).ππ₯ γ I2 =1/4 β«1βγ1/((π₯ + 3/2)^2+ 1/4 ).ππ₯ γ I2 =1/4 β«1βγ1/((π₯ + 3/2)^2+ (1/2)^2 ).ππ₯ γ =1/4 [1/(1/2) tan^(β1)β‘γ(π₯ + 3/2)/(1/2)+πΆ2γ ] =1/4 [2 tan^(β1)β‘γ((2π₯ + 3)/2)/(1/2)+πΆ2γ ] =1/4 [2 γtan^(β1) (2π₯+3)γβ‘γ+πΆ2γ ] It is of form β«1βγππ₯/(π₯^2 + π^2 )=1/π γπ‘ππγ^(β1)β‘γπ₯/π+πΆ2" " γ γ Replacing π₯ by (π₯+3/2) and by 1/2, we get =2/4 tan^(β1)β‘γ(2π₯+3)+πΆ2/4γ =1/2 tan^(β1)β‘γ(2π₯+3)+πΆ3γ Now, putting the value of I1 and I2 in eq. (1) β΄ β«1βγ(π₯+2)/(2π₯^2 + 6π₯ + 5).ππ₯γ =1/4 πππ|2π₯^2+6π₯+5|+πΆ1+1/2 tan^(β1)β‘γ(2π₯+3)+γ πΆ3 =π/π πππ|ππ^π+ππ+π|+π/π γπππγ^(βπ)β‘γ(ππ+π)+γ πͺ (where πΆ3 = πΆ2/4 )