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Integration by parts
Ex 7.6, 3
Ex 7.6, 23 (MCQ)
Example 17
Ex 7.6, 1
Ex 7.6, 2 Important
Ex 7.6, 12
Example 21 Important
Ex 7.6, 21
Ex 7.6, 5 Important
Ex 7.6, 4
Ex 7.6, 6
Ex 7.6, 15
Example 18 Important
Ex 7.6, 14 Important
Ex 7.6, 7 Important
Ex 7.6, 9
Ex 7.6, 8
Ex 7.6, 11
Example 20 Important
Ex 7.6, 13 Important You are here
Ex 7.6, 22 Important
Ex 7.6, 10 Important
Example 38 Important
Integration by parts
Last updated at May 29, 2023 by Teachoo
Ex 7.6, 13 Integrate the function - tan^(−1) 𝑥 ∫1▒〖" " tan^(−1) 𝑥" " 〗 .𝑑𝑥=∫1▒〖(tan^(−1) 𝑥) 1.𝑑𝑥 " " 〗 = tan^(−1) 𝑥∫1▒〖1 .〗 𝑑𝑥−∫1▒(𝑑(tan^(−1)𝑥 )/𝑑𝑥 ∫1▒〖1 .𝑑𝑥〗) 𝑑𝑥 = tan^(−1) 𝑥 (𝑥)−∫1▒1/(1 + 𝑥^2 ) . 𝑥 . 𝑑𝑥 = 𝑥 tan^(−1) 𝑥−∫1▒𝑥/(1 + 𝑥^2 ) . 𝑑𝑥 Now we know that ∫1▒〖𝑓(𝑥) 𝑔(𝑥) 〗 𝑑𝑥=𝑓(𝑥) ∫1▒𝑔(𝑥) 𝑑𝑥−∫1▒(𝑓′(𝑥)∫1▒𝑔(𝑥) 𝑑𝑥) 𝑑𝑥 Putting f(x) = tan–1 x and g(x) = 1 Solving I1 I1 = ∫1▒𝑥/(1 + 𝑥^2 ) . 𝑑𝑥" " Let 1 + 𝑥^2=𝑡 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 0 + 2𝑥=𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡/2𝑥 Our equation becomes I1 = ∫1▒𝑥/(1 + 𝑥^2 ) . 𝑑𝑥" " Putting the value of (1+𝑥^2 ) = t and 𝑑𝑥 = 𝑑𝑡/( 2𝑥) , we get I1 = ∫1▒𝑥/𝑡 . 𝑑𝑡/2𝑥 I1 = 1/2 ∫1▒1/𝑡 . 𝑑𝑡 I1 = 1/2 log〖 |𝑡|〗+𝐶1 I1 = 1/2 log〖 |1+𝑥^2 |〗+𝐶1 Putting the value of I1 in (1) , ∫1▒〖" " tan^(−1) 𝑥" " 〗 .𝑑𝑥=𝑥 tan^(−1) 𝑥−∫1▒𝑥/(1 + 𝑥^2 ) . 𝑑𝑥 =𝑥 tan^(−1) 𝑥−(1/2 〖log 〗|1+𝑥^2 |+𝐶1) =𝑥 tan^(−1) 𝑥−1/2 〖log 〗|1+𝑥^2 |−𝐶1 =𝒙 〖𝒕𝒂𝒏〗^(−𝟏) 𝒙−𝟏/𝟐 〖𝒍𝒐𝒈 〗(𝟏+𝒙^𝟐 )+𝑪