# Ex 7.10, 7 - Chapter 7 Class 12 Integrals (Term 2)

Last updated at Aug. 20, 2021 by Teachoo

Last updated at Aug. 20, 2021 by Teachoo

Transcript

Ex7.10, 7 Evaluate the integrals using substitution −11 𝑑𝑥 𝑥2 + 2𝑥 + 5 we can write −11 𝑑𝑥 𝑥2 + 2𝑥 + 5= −11 𝑑𝑥 𝑥 + 2𝑥 + 1 + 4 = −11 𝑑𝑥 𝑥 + 12 + 22 Putting 𝑥+1=𝑡 Differentiating w.r.t.𝑥 𝑑𝑑𝑥 𝑥+1= 𝑑𝑡𝑑𝑥 1= 𝑑𝑡𝑑𝑥 𝑑𝑥=𝑑𝑡 Hence when 𝑥 varies from – 1 to 1 then 𝑡 varies from 0 to 2 Therefore, −11 𝑑𝑥 𝑥+12 + 22= 02 𝑑𝑡 𝑡2 + 22 = 12 tan−1 𝑡202 = 12 tan−1 22− 12 tan−1 02 = 12 tan−11− 12 tan−10 = 12 × 𝜋4−0 = 𝝅𝟖

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