Solve all your doubts with Teachoo Black (new monthly pack available now!)

Are you in **school**? Do you **love Teachoo?**

We would love to talk to you! Please fill this form so that we can contact you

Ex 7.7

Ex 7.7, 1

Ex 7.7, 2 Important

Ex 7.7, 3

Ex 7.7, 4

Ex 7.7, 5 Important

Ex 7.7, 6

Ex 7.7, 7 Important

Ex 7.7, 8 Important You are here

Ex 7.7, 9

Ex 7.7, 10

Ex 7.7, 11 Important

Ex 7.7, 12 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams

Ex 7.7, 13 (Supplementary NCERT) Deleted for CBSE Board 2023 Exams

Ex 7.7, 14 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams

Chapter 7 Class 12 Integrals

Serial order wise

Last updated at Dec. 20, 2019 by Teachoo

Ex 7.7, 8 β(π₯2+3π₯) β«1βγβ(π₯^2+3π₯) ππ₯γ =β«1βγβ(π₯^2+2(3/2)(π₯) ) ππ₯γ =β«1βγβ(π₯^2+2(3/2)(π₯)+(3/2)^2β(3/2)^2 ) ππ₯γ =β«1βγβ((π₯+3/2)^2β(3/2)^2 ) ππ₯γ It is of the form β«1βγβ(π₯^2βπ^2 ) ππ₯=π₯/2 β(π₯^2βπ^2 )βπ^2/2 πππ|π₯+β(π₯^2βπ^2 )|+πΆγ β΄ Replacing π₯ by π₯+3/2 and a by 3/2 , we get =(π₯ + 3/2)/2 β((π₯+3/2)^2β(3/2)^2 )β|π₯+3/2+β((π₯+π₯/2)^2β(3/2)^2 )|+πΆ =(2π₯ + 3)/4 β((π₯+3/2)^2β9/4) β 9/8 πππ|π₯+3/2 β((π₯+3/2)^2β9/4)|+πΆ =(2π₯ + 3)/4 β(π₯^2+9/4+2(π₯)(3/2)β9/4) β 9/8 πππ|π₯+3/2+β(π₯^2+9/4+2 (π₯)(3/2)β9/4)|+πΆ =(2π₯ + 3)/4 β(π₯^2+2π₯(3/2) )β 9/8 πππ|π₯+3/2+β(π₯^2+2π₯(3/2) )|+πΆ =(ππ + π)/π β(π^π+ππ)β π/π πππ|π+π/π+β(π^π+ππ)|+πͺ