1. Class 12
2. Important Question for exams Class 12
3. Chapter 7 Class 12 Integrals

Transcript

Example 9 - Chapter 7 Class 12 Integration - NCERT Solution Find the following integrals: (i) β« ππ₯/(π₯^2 β 6π₯ + 13) (ii) β« ππ₯/(3π₯^2 β 13π₯ + 10) (iii) β« ππ₯/(β(5π₯^2 β 2π₯)) Let's solve them part by part (i) β« ππ₯/(π₯^2β 6π₯ + 13) β« ππ₯/(π₯^2β 6π₯ + 13) =β« ππ₯/(π₯^2β 2 Γ 3 Γ π₯ + 13) = β« ππ₯/((π₯^2 β 2 . 3 π₯ + 3^2 ) + 13 β 3^2 )("Adding and subtracting " (3)^2 ) = β« ππ₯/((π₯ β 3)^2 + 13 β 9)("Using " π^2β2ππ+π^2=(πβπ)^2 ) = β« ππ₯/((π₯ β 3)^2 + 4) = β« ππ₯/((π₯ β 3)^2 + 2^2 ) It is of form ππ₯/(π₯^2 + π^2 )=1/π π‘ππ ^(β1)β‘ π₯/π +πΆ Replacing π₯ with (π₯β3) and π with 2 =π/π πππ ^(βπ)β‘ (π β π)/π +πͺ (ii) β« ππ₯/(3π₯^2 β 13π₯ + 10) β« ππ₯/( 3π₯^2 β 13π₯ + 10) Solving denominator 3π₯^2 + 13π₯ β 10 =3(π₯^2+13/3 π₯ β10/3) =3(π₯^2+2. π₯Γ 13/6 β10/3) Adding and subtracting (13/6)^2 =3(π₯^2+2. π₯Γ 13/6+(13/6)^2β10/3β(13/6)^2 ) =3((π₯+13/6)^2β10/3β(169/36)) =3((π₯+13/6)^2β(10/3 +169/36)) =3((π₯+13/6)^2β((120 + 69)/36 )) =3((π₯+13/6)^2β189/36) =3((π₯+13/6)^2β(17/6)^2 ) Hence, our equation becomes β« ππ₯/( 3π₯ ^2β13π₯ + 10) = 1/3 β« ππ₯/((π₯ + 13/6)^2β (17/6)^2 ) It is of form β« ππ₯/(π₯^2 β π^2 )=1/2π πππ|(π₯ β π)/(π₯ + π)|+πΆ1 Replacing π₯ by (π₯+13/6)πππ π ππ¦ 17/6, = 1/3 Γ 1/2(17/6) Γlogβ‘|(π₯ + 13/6 β 17/6)/(π₯+ 13/6 + 17/6)| + C = 1/3 Γ 6/2(17) Γlogβ‘|((6π₯ + 13 β 17)/6)/((6π₯ +13 + 17)/6)| + C = 1/17 logβ‘|(6π₯ β 4)/(6π₯ + 30)| + C = 1/17 logβ‘|(2(3π₯ β 2))/(6(π₯ + 5))|+ C = 1/17 logβ‘|( (3π₯ β 2))/(3(π₯ + 5))|+ C = 1/17 logβ‘|( (3π₯ β 2))/((π₯ + 5))|β1/17 logβ‘3 + C(As log π/π = log a β log b) = π/ππ πππβ‘|( (ππ β π))/((π + π))|+ C1 Example 9 Find the following integrals: (iii) β« ππ₯/(β(5π₯^2 β 2π₯) ) β« ππ₯/(β(5π₯^2 β 2π₯) ) = β« ππ₯/(β(5(π₯^2 β 2/5 π₯) ) ) (Taking 5 common) = β« ππ₯/(β(5(π₯^2 β 2(π₯)(1/5)) ) ) = β« ππ₯/(β(5(π₯^2 β 2(π₯)(1/5) + (1/5)^2β (1/5)^2 ) ) ) [Adding and subtracting (1/5)^2] = β« ππ₯/(β(5[(π₯ β 1/5)^2β(1/5)^2 ] ) ) [using π^2β2ππ+π^2=(πβπ)^2] = β« ππ₯/(β5 β((π₯ β 1/5)^2β(1/5)^2 ))[usingβ(π.π)=βπ βπ] It is of form β« ππ₯/(β(π₯^2βπ^2 ) )=πππ|π₯+β(π₯^2βπ^2 )|+πΆ1 Replacing π₯ by (π₯β1/5)πππ π ππ¦ 1/5, =1/β5 [πππ|π₯β1/5+β((π₯β1/5)^2β(1/5)^2 )|+πΆ1] =1/β5 πππ|π₯β1/5+β((π₯β1/5)^2β(1/5)^2 )|+πΆ1/β5 =1/β5 πππ|π₯β1/5+β(π₯^2+(1/5)^2β2(π₯)(1/5)β(1/5)^2 )|+πΆ("c = " πΆ1/β5) =π/βπ πππ|πβπ/π+β(π^πβππ/π)|+πͺ

Chapter 7 Class 12 Integrals

Class 12
Important Question for exams Class 12