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Ex 7.7, 12 (Supplementary NCERT) - Integrate x root(x + x2)

Ex 7.7, 12 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 2
Ex 7.7, 12 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 3 Ex 7.7, 12 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 4 Ex 7.7, 12 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 5 Ex 7.7, 12 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 6 Ex 7.7, 12 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 7

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Ex 7.7, 12 (Supplementary NCERT) ( + ^2 ) ( + ^2 ) Step 1: It can be written in the form x = A [ / ( + ^2 )]+ B x = A [1+2 ]+ B x = (2A) x + A +B x = (2A) x / = 2A 1 = 2A A = 1/20 = A + B B = A B = ( 1)/2 Thus, x = A [1 + 2x] + B x = 1/2 [1 + 2x] 1/2 Step 2: Integrating the function w.r.t. x 1 ( + ^2 ) = 1 [1/2 [1+2 ] 1/2] ( + ^2 ) = 1 1/2 [1+2 ] ( + ^2 ) 1 1/2 ( + ^2 ) = 1/2 1 (1+2 ) ( + ^2 ) 1/2 1 ( + ^2 ) Solving _ _1= 1/2 1 (1+2 ) ( + ^2 ) . Let + ^2= Diff. both sides w.r.t.x 1 + 2 = / dx = /(1 + 2 ) Thus, our equation becomes As I_1= 1/2 1 (1+2 ) ( + ^2 ) . Putting the value of ( + ^2) and d I_1= 1/2 1 (1+2 ) . /((1 + 2 ) ) I_1= 1/2 1 I_1= 1/2 1 ( )^(1/2) I_1= 1/2 ^( 1/2 ^(+ 1) )/((1/2 + 1) ) + C_1 I_1= 1/2 ^((1 + 2)/2)/(((1 + 2)/2 ) ) + C_1 I_1= 1/2 ^(3/2)/(3/2) + C_1 I_1= ^(3/2)/3 + C_1 I_1= ( ^2 + ) ^(3/2)/3 + C_1 Solving _ I_2= 1/2 1 ( + ^2 ) I_2 = 1/2 1 ( ^2+ ) I_2 = 1/2 1 ( ^2+2( )(1/2) ) I_2 = 1/2 1 ( ^2+2( )(1/2)+(1/2)^2 (1/2)^2 ) I_2 = 1/2 1 (( +1/2)^2 (1/2)^2 ) It is of form 1 ( ^2 ^2 ) = /2 ( ^2 ^2 ) ^2/2 log | + ( ^2 ^2 )|+C_2 Replacing x by (x + 1/2 ) and a by 1/2, we get I_2 = 1/2 [( + 1/2)/2 (( +1/2)^2 (1/2)^2 ) (1/2)^2/2 log | +1/2+ (( +1/2)^2 (1/2)^2 )|+ C_2 ] I_2 = 1/2 [((2 +1)/2)/2 (( +1/2)^2 (1/2)^2 ) (1/2)^2/2 log| +1/2+ (( +1/2)^2 (1/2)^2 )|+ C_2 ] I_2 = 1/2 [(2 +1)/2.2 ( ^2 (1/2)^2+2( )(1/2) (1/2)^2 ) (1/4)/2 log | +1/2+ (( +1/2)^2+2( )(1/2) ) + C_2 ] I_2 = 1/2 [(2 +1)/4 ( ^2+ ) 1/(4 . 2 ) log | +1/2+ (( )^2+2( )(1/2) ) + C_2 ] I_2 = 1/2 [(2 +1)/4 ( ^2+ ) 1/8 log | +1/2+ (( )^2+ )|+ C_2 ] I_2 = ((2 + 1))/8 ( ^2+ ) 1/16 log | +1/2+ (( )^2+ )|+C_3 Putting the value of I_1 and I_2 in (1) , we get 1 ( + ^2 ) = 1/2 1 (1+2 ) ( + ^2 ) 1/2 1 ( + ^2 ) = ( + ^2) ^(3/2)/3+C_1 ((2 + 1) ( ^2 + ))/8+1/16 log | +1/2+ ( ^2+ )| C_3 = ( + ^ ) ^( / )/ (( + ) ( ^ + ))/ + / | + / + ( ^ + )|+

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.