![Example 40 - Evaluate [log (log x) + 1 / (log x)2 ] dx - Examples](https://d1avenlh0i1xmr.cloudfront.net/35f3b842-e093-4f51-b6f3-4b982d60fb9a/slide26.jpg)
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Example 38 - Chapter 7 Class 12 Integrals
Last updated at May 29, 2023 by Teachoo
Examples
Example 1 (i)
Example 1 (ii)
Example 1 (iii)
Example 2 (i)
Example 2 (ii)
Example 2 (iii) Important
Example 3 (i)
Example 3 (ii) Important
Example 3 (iii)
Example 4
Example 5 (i)
Example 5 (ii)
Example 5 (iii) Important
Example 5 (iv) Important
Example 6 (i)
Example 6 (ii) Important
Example 6 (iii) Important
Example 7 (i)
Example 7 (ii) Important
Example 7 (iii)
Example 8 (i)
Example 8 (ii) Important
Example 9 (i)
Example 9 (ii) Important
Example 9 (iii) Important
Example 10 (i)
Example 10 (ii) Important
Example 11
Example 12
Example 13 Important
Example 14
Example 15 Important
Example 16 Important
Example 17
Example 18 Important
Example 19
Example 20 Important
Example 21 Important
Example 22 Important
Example 23
Example 24
Example 25 (i)
Example 25 (ii) Important
Example 25 (iii)
Example 25 (iv) Important
Example 26
Example 27
Example 28 Important
Example 29
Example 30
Example 31
Example 32 Important
Example 33 Important
Example 34 Important
Example 35
Example 36 Important
Example 37 Important
Example 38 Important You are here
Example 39 Important
Example 40 Important
Example 41 Important
Example 42 Important
Question 1 Important Deleted for CBSE Board 2024 Exams
Question 2 Deleted for CBSE Board 2024 Exams
Question 3 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams
Chapter 7 Class 12 Integrals
Serial order wise
Transcript
Example 38 Evaluate β«1β[logβ‘(logβ‘π₯ )+1/(logβ‘π₯ )^2 ] ππ₯ Let I1 =β«1β[πππ(logβ‘π₯ )+1/(logβ‘π₯ )^2 ]ππ₯ I1 = β«1βγπππ(logβ‘π₯ )ππ₯+β«1βγ1/(logβ‘π₯ )^2 .ππ₯γγ Solving ππ I2 =β«1βπππ(logβ‘π₯ )ππ₯ I2 =β«1βγπππ(logβ‘π₯ ).1 ππ₯γ Using by parts β«1βγπ(π₯) πβ‘(π₯) γ ππ₯=π(π₯) β«1βπ(π₯) ππ₯ββ«1β(π^β² (π₯) β«1βπ(π₯) ππ₯) ππ₯ Putting f(x) = log (log x) and g(x) = 1 I2=πππ(ππππ₯) β«1βγ1.ππ₯ββ«1β[π[πππ[ππππ₯]]/ππ₯ β«1βγ1.ππ₯γ] γ ππ₯ I2=log(ππππ₯)(π₯)ββ«1β[1/logβ‘π₯ .π(ππππ₯)/ππ₯ β«1βγ1.ππ₯γ]ππ₯ I2=π₯ πππ(ππππ₯)ββ«1βγ1/logβ‘π₯ .1/π₯ .π₯ ππ₯γ I2=π₯ πππ(ππππ₯)ββ«1βγ1/logβ‘π₯ .1/π₯ .π₯ ππ₯γ I2=π₯ πππ(ππππ₯)ββ«1βγ1/logβ‘π₯ ππ₯γ Solving ππ I3=β«1βγ1/logβ‘π₯ .ππ₯γ I3=β«1βγ1/logβ‘π₯ .1 ππ₯γ I3=1/ππππ₯ β«1βγ1.ππ₯ββ«1β[π(1/ππππ₯)/ππ₯ β«1βγ1.ππ₯γ] γ ππ₯ I3=1/ππππ₯.π₯ββ«1βγ(β1)/(ππππ₯)^2 (1/π₯).π₯ ππ₯γ Using by parts β«1βγπ(π₯) πβ‘(π₯) γ ππ₯=π(π₯) β«1βπ(π₯) ππ₯ββ«1β(π^β² (π₯) β«1βπ(π₯) ππ₯) ππ₯ Putting f(x) = 1/πππβ‘π₯ and g(x) = 1 I3=π₯/logβ‘π₯ +β«1βγ1/(logβ‘π₯ )^2 ππ₯γ Putting the value of I3 in I2 , we get I2=π₯ πππ(logβ‘π₯ )βπΌ3 I2=π₯ πππ(logβ‘π₯ )β[π₯/logβ‘π₯ +β«1β1/(logβ‘π₯ )^2 ππ₯] I2=π₯ πππ(logβ‘π₯ )βπ₯/logβ‘π₯ ββ«1βγ1/(logβ‘π₯ )^2 ππ₯γ Now, Putting the value of πΌ2 in πΌ1 , we get I1=I2+β«1βγ1/(logβ‘π₯ )^2 ππ₯γ I1=π₯ πππ(logβ‘π₯ )βπ₯/logβ‘π₯ ββ«1βγ1/(logβ‘π₯ )^2 ππ₯γ+β«1βγ1/(logβ‘π₯ )^2 ππ₯γ β΄ I1=π πππ(πππβ‘π )βπ/πππβ‘π + C
