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Integration by parts
Ex 7.6, 3
Ex 7.6, 23 (MCQ)
Example 17
Ex 7.6, 1
Ex 7.6, 2 Important
Ex 7.6, 12
Example 21 Important
Ex 7.6, 21
Ex 7.6, 5 Important
Ex 7.6, 4
Ex 7.6, 6
Ex 7.6, 15
Example 18 Important
Ex 7.6, 14 Important
Ex 7.6, 7 Important
Ex 7.6, 9
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Example 20 Important
Ex 7.6, 13 Important
Ex 7.6, 22 Important
Ex 7.6, 10 Important
Example 40 Important You are here
Integration by parts
Last updated at Dec. 23, 2019 by Teachoo
Example 40 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