Check sibling questions

Ex 7.9, 19 - Direct Integrate 6x + 3 / x2 + 4 dx from 0 to 2

Ex 7.9, 19 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.9, 19 - Chapter 7 Class 12 Integrals - Part 3 Ex 7.9, 19 - Chapter 7 Class 12 Integrals - Part 4

Get live Maths 1-on-1 Classs - Class 6 to 12


Transcript

Ex 7.9, 19 ∫_0^2▒(6𝑥 + 3)/(𝑥^2 + 4) 𝑑𝑥 Let F(𝑥)=∫1▒〖(6𝑥 + 3)/(𝑥^2 + 4) 𝑑𝑥〗 =∫1▒〖6𝑥/(𝑥^2 + 4) 𝑑𝑥+∫1▒〖3/(𝑥^2 + 4) 𝑑𝑥 〗〗 Solving 𝑰𝟏 𝐼1=∫1▒〖6𝑥/(𝑥^2 + 4) 𝑑𝑥〗 Put 𝑥^2 + 4=𝑡 Differentiating w.r.t.𝑥 𝑑/𝑑𝑥 (𝑥^2+4)=𝑑𝑡/𝑑𝑥 2𝑥+0=𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡/2𝑥 Therefore, ∫1▒〖6𝑥/(𝑥^2 + 4) 𝑑𝑥=∫1▒〖6𝑥/𝑡 𝑑𝑡/2𝑥〗〗 =∫1▒〖3/𝑡 𝑑𝑡〗 =3 𝑙𝑜𝑔|𝑡| =3 𝑙𝑜𝑔|𝑥^2+4| (∵𝑡=𝑥^2+4) Solving 𝑰𝟐 𝐼2=∫1▒〖3/(𝑥^2 + 4) 𝑑𝑥〗 =3∫1▒1/(𝑥^2+4) 𝑑𝑥 =3∫1▒1/(𝑥^2 + 2^2 ) 𝑑𝑥 =3 × 1/2 tan^(−1)⁡〖𝑥/2〗 =3/2 tan^(−1)⁡〖𝑥/2〗 Therefore F(𝑥)= 𝐼1+𝐼2 F(𝑥)=3𝑙𝑜𝑔|𝑥^2+4|+3/2 tan^(−1)⁡〖𝑥/2〗 ( As ∫1▒〖𝑑𝑥/(𝑥^2 + 𝑎^2 )=1/𝑎 tan^(−1)⁡〖𝑥/𝑎〗 〗) Now, ∫_0^2▒〖(6𝑥 + 3)/(𝑥^2 + 4) 𝑑𝑥=𝐹(2)−𝐹(0) 〗 =3𝑙𝑜𝑔|2^2+4|+3/2 tan^(−1)⁡〖2/2−3𝑙𝑜𝑔|0+4|−3/2 tan^(−1)⁡(0/2) 〗 =3𝑙𝑜𝑔|4+4|+3/2 tan^(−1)⁡〖1−3𝑙𝑜𝑔|4|−3/2 × 0〗 =3𝑙𝑜𝑔|8|−3𝑙𝑜𝑔|4|+3/2 𝜋/4 =3(𝑙𝑜𝑔|8|−𝑙𝑜𝑔|4|)+3𝜋/8 =3𝑙𝑜𝑔|8/4|+3𝜋/8 =𝟑 𝐥𝐨𝐠⁡〖𝟐+𝟑𝝅/𝟖〗 (As log A – log B = log A/B )

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

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.