Examples

Chapter 7 Class 12 Integrals
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Example 27 Evaluate the following integrals: (iii) β«_1^2β(π₯ ππ₯)/((π₯ + 1) (π₯ + 2) ) ππ₯ πΉ(π₯)=β«1β(π₯ ππ₯)/(π₯ + 1)(π₯ + 2) We can write the integrate as : π₯/(π₯ + 1)(π₯ + 2) =A/(π₯ + 1)+B/(π₯ + 2) π₯/(π₯ + 1)(π₯ + 2) =(A(π₯ + 2) + B(π₯ + 1))/(π₯ + 1)(π₯ + 2) Canceling denominators π₯=A(π₯+2)+B(π₯+1) Putting π=βπ β2=A(2+2)+B(2+1) β2=A Γ0+B(β1) β2=βB B=2 Putting π=βπ β1=A(β1+2)+B(β1+1) β1=A(1)+B(0) β1=A A=β1 Therefore π₯/(π₯ + 1)(π₯ + 2) =(β1)/(π₯ + 1)+2/(π₯ + 2) Integrating w.r.t.π₯ β«1βγπ₯/(π₯+1)(π₯+2) =β«1βγ(β1)/((π₯+1) ) ππ₯γ+β«1βγ2/(π₯+2) ππ₯γγ =βπππ|π₯+1|+2πππ|π₯+2| =βπππ|π₯+1|+πππ|π₯+2|^2 =πππ|π₯+2|^2βπππ|π₯+1| =πππ|(π₯ + 2)^2/(π₯ + 1)| Hence πΉ(π₯)=πππ|(π₯ + 2)^2/(π₯ + 1)| Now, β«_1^2βγπ₯/(π₯ + 1)(π₯ + 2) ππ₯γ=πΉ(2)βπΉ(1) β«_1^2βγπ₯/(π₯ + 1)(π₯ + 2) ππ₯γ=πππ|(2 + 2)^2/(2 + 1)|βπππ|(1 + 2)^2/(1 + 1)| (π logβ‘γπ=logβ‘γπ^π γ γ ) (logβ‘Aβlogβ‘B=πππ A/π΅) =πππ|(4)^2/3|βπππ|(3)^2/2| =πππ|(4^2/3)/(3^2/2)| =πππ|4^2/3 Γ 2/3^2 | =πππ|16/3 Γ 2/9| =πππ|32/27 | =π₯π¨π β‘(ππ/ππ) (logβ‘Aβlogβ‘B=πππ A/π΅) (logβ‘Aβlogβ‘B=πππ A/π΅)