2 / (x + 1)2 (x + 3) dx

Basic Integration Example 07 (Changing Base of an Exponential Function)
Basic Integration Example 07 (Changing Base of an Exponential Function)

Examples

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Question 1 Important Deleted for CBSE Board 2024 Exams

Question 2 Deleted for CBSE Board 2024 Exams

Question 3 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams

Last updated at May 29, 2023 by Teachoo

Example 13 Find ∫1▒(3𝑥 −2)/((𝑥 + 1)^2 (𝑥 + 3) ) 𝑑𝑥 We can write Integral as (3𝑥 − 2)/((𝑥 + 1)^2 (𝑥 + 3) )=𝐴/(𝑥 + 1) + 𝐵/(𝑥 + 1)^2 + 𝐶/((𝑥 + 3) ) (3𝑥 − 2)/((𝑥 + 1)^2 (𝑥 + 3) )=(𝐴(𝑥 + 1)(𝑥 + 3) + 𝐵(𝑥 + 3) + 𝐶(𝑥 + 1)^2)/((𝑥 + 1)^2 (𝑥 + 3) ) Cancelling denominator 3𝑥 −2=𝐴(𝑥+1)(𝑥+3)+𝐵(𝑥+3)+𝐶(𝑥+1)^2 Putting x = −1 3(−1) −2=𝐴(−1+1)(−1+3)+𝐵(−1+3)+𝐶(−1+1)^2 −3−2=𝐴×0+𝐵×2+𝐶×(0)^2 −5=𝐵×2 𝐵=(− 5)/2 Putting x = − 3 3𝑥 −2=𝐴(𝑥+1)(𝑥+3)+𝐵(𝑥+3)+𝐶(𝑥+1)^2 3(−3)−2=𝐴(−3+1)(−3+3)+𝐵(−3+3)+𝐶(−3+1)^2 −9−2=𝐴×0+𝐵×0+𝐶×(−2)^2 −11=0+0+𝐶(4) −11=4𝐶 (−11)/4 =𝐶 𝐶 =(−11)/4 Putting x = 0 3𝑥 −2=𝐴(𝑥+1)(𝑥+3)+𝐵(𝑥+3)+𝐶(𝑥+1)^2 3(0) − 2 = A(1) (3) + B(3) + C 〖”(1)” 〗^2 −2 = 3A + 3B + C Putting value of B & C −2 = 3A + 3((−5)/2) + ((−11)/4) −2 = 3A − 15/2−11/4 −2 = 3A + (−30 − 11)/4 −8 = 12A − 41 41 − 8 = 12A 33/12 = A 11/4 = A Hence, we can write (3𝑥 − 2)/((𝑥 + 1)^2 (𝑥 + 3) )=𝐴/(𝑥 + 1) + 𝐵/(𝑥 + 1)^2 + 𝐶/((𝑥 + 3) ) (3𝑥 − 2)/((𝑥 + 1)^2 (𝑥 + 3) )=11/4(𝑥 + 1) − 5/〖2(𝑥 + 1)〗^2 − 11/4(𝑥 + 3) Therefore ∫1▒(3𝑥 − 2)/((𝑥 + 1)^2 (𝑥 + 3) ) 𝑑𝑥 =∫1▒11/4(𝑥 + 1) 𝑑𝑥−∫1▒5/〖2(𝑥 + 1)〗^2 𝑑𝑥−∫1▒11/4(𝑥 + 3) 𝑑𝑥 =11/4 log|𝑥+1|−5/2×((−1))/((𝑥 + 1) ) − 11/4 log|𝑥+3|+𝐶 =11/4 (log|𝑥+1|−log|𝑥+3| )+5/(2 (𝑥 + 1) )+𝐶 =𝟏𝟏/𝟒 𝒍𝒐𝒈|(𝒙 + 𝟏)/(𝒙 + 𝟑)| + 𝟓/(𝟐 (𝒙 + 𝟏) )+𝑪

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