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Ex 7.7

Ex 7.7, 1

Ex 7.7, 2 Important

Ex 7.7, 3

Ex 7.7, 4

Ex 7.7, 5 Important

Ex 7.7, 6

Ex 7.7, 7 Important

Ex 7.7, 8 Important

Ex 7.7, 9

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Ex 7.7, 11 Important

Ex 7.7, 12 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams

Ex 7.7, 13 (Supplementary NCERT) Deleted for CBSE Board 2024 Exams You are here

Ex 7.7, 14 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams

Last updated at May 29, 2023 by Teachoo

Ex 7.7, 13 (Supplementary NCERT) 𝑥+1 √(2𝑥^2+3) 𝑥+1 √( 2𝑥^2+3) Integrating the function w.r.t.x ∫1▒〖(𝑥+1) √(〖2𝑥〗^2+3)〗 𝑑𝑥 = ∫1▒〖[𝑥√(〖2𝑥〗^2+3)+1√(〖2𝑥〗^2+3)] 𝑑𝑥〗 = ∫1▒〖𝑥 √(〖2𝑥〗^2+3) 𝑑𝑥+∫1▒〖√(〖2𝑥〗^2+3 ) 𝑑𝑥〗〗 Solving 𝐈_𝟏 𝐈_𝟏=∫1▒〖𝑥 √(〖2𝑥〗^2+3)〗 𝑑𝑥 Let 〖2𝑥〗^2+3=𝑡 Diff. both sides w.r.t.x 4x + 0 = 𝑑𝑡/𝑑𝑥 dx = 𝑑𝑡/4𝑥 Now, equation becomes I_1 = ∫1▒〖𝑥 √(〖2𝑥〗^2+3)〗 𝑑𝑥 Putting the value of (〖2𝑥〗^2+3) and dx I_1 = ∫1▒〖𝑥 √𝑡〗 . 𝑑𝑥 I_1 = ∫1▒〖𝑥 √𝑡〗 𝑑𝑡/4𝑥 I_1 = 1/4 ∫1▒〖 √𝑡 〗. 𝑑𝑡 I_1 = 1/4 ∫1▒〖 𝑡^(1/2) 〗 . 𝑑𝑡 I_1 = 1/4 [〖𝑡 〗^(1/2 + 1)/((1/2 + 1) )]+C_1 I_1 = 1/4 [〖𝑡 〗^(3/2 )/(3/2)]+C_1 I_1 = 〖𝑡 〗^(3/2 )/6+C_1 I_1 = (〖2𝑥〗^2+3)^(3/2)/6+C_1 Solving 𝑰_𝟐 I_2 = ∫1▒√(〖2𝑥〗^2+3) 𝑑𝑥 I_2 = ∫1▒√(2 (〖 𝑥〗^2+3/2) ) 𝑑𝑥 I_2 = ∫1▒〖√2 √( 〖 𝑥〗^2+3/2)〗 𝑑𝑥 I_2 =√2 ∫1▒√(〖 𝑥〗^2+3/2) 𝑑𝑥 I_2 =√2 ∫1▒√(〖 𝑥〗^2+(√(3/2))^2 ) 𝑑𝑥 It is of form ∫1▒〖√(𝑥^2+𝑎^2 ) 〗 𝑑𝑥=1/2 𝑥 √(𝑥^2+𝑎^2 )+𝑎^2/2 log|𝑥+√(𝑥^2+𝑎^2 )|+C_2 Replacing a by √(3/2), we get I_2 =√2 [𝑥/2 √(〖 𝑥〗^2+(√(3/2))^2 )]+√2 〖((√(3/2 )))/2〗^2 log|𝑥+√(〖 𝑥〗^2+(√(3/2))^2 )|+C_2 I_2 =√2 [𝑥/2 √(〖 𝑥〗^2+3/2)]+√2 (3/2)/2 log|𝑥+√(〖 𝑥〗^2+3/2)|+C_2 I_2 =(√2 𝑥)/2.√(〖 2𝑥〗^2+3)/√2+(3√2)/4 log|𝑥+√(〖 𝑥〗^2+3/2)|+〖√2 C〗_2 I_2 =𝑥/2.√(〖 2𝑥〗^2+3)+(3√2)/4 log|𝑥+√(〖 𝑥〗^2+3/2)|+〖 C〗_3 Now, put the value of I_1 and I_2 in eq. (1), we get ∫1▒〖(𝑥+1) √(〖2𝑥〗^2+3)〗 𝑑𝑥=∫1▒𝑥 √(〖2𝑥〗^2+3) 𝑑𝑥+∫1▒√(〖2𝑥〗^2+3) 𝑑𝑥 = (〖2𝑥〗^2+3)^(3/2)/6+〖 C〗_1+𝑥/2 √(〖2𝑥〗^2+3)+(3√2)/4 log|𝑥+√(〖 𝑥〗^2+3/2)|+〖 C〗_3 = (〖𝟐𝒙〗^𝟐+𝟑)^(𝟑/𝟐)/𝟔+𝒙/𝟐 √(〖𝟐𝒙〗^𝟐+𝟑)+(𝟑√𝟐)/𝟒 𝒍𝒐𝒈|𝒙+√(〖 𝒙〗^𝟐+𝟑/𝟐)|+𝐂