Examples

Chapter 7 Class 12 Integrals
Serial order wise

### Transcript

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

Made by

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.