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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Ex 7.8, 14 ∫_0^1▒(2𝑥 + 3)/(5𝑥2 + 1) dx Let F(𝒙)=∫1▒(2𝑥 + 3)/(5𝑥^2 + 1) 𝑑𝑥 =∫1▒2𝑥/(5𝑥^2 + 1) 𝑑𝑥+∫1▒3/(5𝑥^2 + 1) 𝑑𝑥 Solving ∫1▒𝟐𝒙/(𝟓𝒙^𝟐 + 𝟏) 𝒅𝒙 Put 𝑥^2=𝑡 Differentiating w.r.t.𝑥 2𝑥=𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡/2𝑥 Hence ∫1▒〖(2𝑥 )/(5𝑥^2 + 1) 𝑑𝑥=∫1▒〖2𝑥/(5𝑡+1) 𝑑𝑡/2𝑥〗〗 =∫1▒𝑑𝑡/(5𝑡+1) =1/5 𝑙𝑜𝑔|5𝑡+1| =𝟏/𝟓 𝒍𝒐𝒈|𝟓𝒙^𝟐+𝟏| Integrating ∫1▒〖𝟑/(𝟓𝒙^𝟐+𝟏) 𝒅𝒙〗 ∫1▒〖3/(5𝑥^2+1) 𝑑𝑥〗 =3∫1▒𝑑𝑥/(5𝑥^2+1) =3/5 ∫1▒𝑑𝑥/(𝑥^2 + 1/5) =3/5 ∫1▒𝑑𝑥/(𝑥^2 +(1/√5)^2 ) =3/5×(1/1)/√5 tan^(−1) ((𝑥/1)/√5) =3/5×√5 〖𝑡𝑎𝑛〗^(−1) (√5 𝑥) =𝟑/𝟓 〖𝒕𝒂𝒏〗^(−𝟏) (√𝟓 𝒙) Hence, F(𝒙)=∫1▒2𝑥/(5𝑥^2 + 1) 𝑑𝑥+∫1▒3/(5𝑥^2 + 1) 𝑑𝑥 =𝟏/𝟓 𝒍𝒐𝒈|𝟓𝒙^𝟐+𝟏|+𝟑/√𝟓 〖𝒕𝒂𝒏〗^(−𝟏) (√𝟓 𝒙) Now, ∫_0^1▒〖(2𝑥 + 3)/(〖5𝑥〗^2+ 1) 𝑑x=𝐅(𝟏)−𝐅(𝟎)〗 =1/5 𝑙𝑜𝑔|5〖×1〗^2+1|+3/√5 〖𝑡𝑎𝑛〗^(−1) (√5 𝑥) −[1/5 𝑙𝑜𝑔|5×0+1|+3/√5 〖𝑡𝑎𝑛〗^(−1) (√5×0)] =1/5 |6|+3/√5 〖𝑡𝑎𝑛〗^(−1) √5−1/5 𝑙𝑜𝑔1+3/5 〖𝑡𝑎𝑛〗^(−1) 0 =1/5 𝑙𝑜𝑔 6+3/√5 〖𝑡𝑎𝑛〗^(−1) √5−1/5×0+3/5×0 =𝟏/𝟓 𝒍𝒐𝒈 𝟔+𝟑/√𝟓 〖𝒕𝒂𝒏〗^(−𝟏) √𝟓

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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.