Examples

Chapter 7 Class 12 Integrals
Serial order wise

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Example 13 Find β«1β(3π₯ β2)/((π₯ + 1)^2 (π₯ + 3) ) ππ₯ We can write Integral as (3π₯ β 2)/((π₯ + 1)^2 (π₯ + 3) )=π΄/(π₯ + 1) + π΅/(π₯ + 1)^2 + πΆ/((π₯ + 3) ) (3π₯ β 2)/((π₯ + 1)^2 (π₯ + 3) )=(π΄(π₯ + 1)(π₯ + 3) + π΅(π₯ + 3) + πΆ(π₯ + 1)^2)/((π₯ + 1)^2 (π₯ + 3) ) Cancelling denominator 3π₯ β2=π΄(π₯+1)(π₯+3)+π΅(π₯+3)+πΆ(π₯+1)^2 Putting x = β1 3(β1) β2=π΄(β1+1)(β1+3)+π΅(β1+3)+πΆ(β1+1)^2 β3β2=π΄Γ0+π΅Γ2+πΆΓ(0)^2 β5=π΅Γ2 π΅=(β 5)/2 Putting x = β 3 3π₯ β2=π΄(π₯+1)(π₯+3)+π΅(π₯+3)+πΆ(π₯+1)^2 3(β3)β2=π΄(β3+1)(β3+3)+π΅(β3+3)+πΆ(β3+1)^2 β9β2=π΄Γ0+π΅Γ0+πΆΓ(β2)^2 β11=0+0+πΆ(4) β11=4πΆ (β11)/4 =πΆ πΆ =(β11)/4 Putting x = 0 3π₯ β2=π΄(π₯+1)(π₯+3)+π΅(π₯+3)+πΆ(π₯+1)^2 3(0) β 2 = A(1) (3) + B(3) + C γ"(1)" γ^2 β2 = 3A + 3B + C Putting value of B & C β2 = 3A + 3((β5)/2) + ((β11)/4) β2 = 3A β 15/2β11/4 β2 = 3A + (β30 β 11)/4 β8 = 12A β 41 41 β 8 = 12A 33/12 = A 11/4 = A Hence, we can write (3π₯ β 2)/((π₯ + 1)^2 (π₯ + 3) )=π΄/(π₯ + 1) + π΅/(π₯ + 1)^2 + πΆ/((π₯ + 3) ) (3π₯ β 2)/((π₯ + 1)^2 (π₯ + 3) )=11/4(π₯ + 1) β 5/γ2(π₯ + 1)γ^2 β 11/4(π₯ + 3) Therefore β«1β(3π₯ β 2)/((π₯ + 1)^2 (π₯ + 3) ) ππ₯ =β«1β11/4(π₯ + 1) ππ₯ββ«1β5/γ2(π₯ + 1)γ^2 ππ₯ββ«1β11/4(π₯ + 3) ππ₯ =11/4 logβ‘|π₯+1|β5/2Γ((β1))/((π₯ + 1) ) β 11/4 logβ‘|π₯+3|+πΆ =11/4 (logβ‘|π₯+1|βlogβ‘|π₯+3| )+5/(2 (π₯ + 1) )+πΆ =ππ/π πππβ‘|(π + π)/(π + π)| + π/(π (π + π) )+πͺ