Integration by parts

Chapter 7 Class 12 Integrals
Concept wise

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Example 40 Evaluate β«1β[logβ‘(logβ‘π₯ )+1/(logβ‘π₯ )^2 ] ππ₯ Let I1 =β«1β[πππ(logβ‘π₯ )+1/(logβ‘π₯ )^2 ]ππ₯ I1 = β«1βγπππ(logβ‘π₯ )ππ₯+β«1βγ1/(logβ‘π₯ )^2 .ππ₯γγ Solving ππ I2 =β«1βπππ(logβ‘π₯ )ππ₯ I2 =β«1βγπππ(logβ‘π₯ ).1 ππ₯γ Using by parts β«1βγπ(π₯) πβ‘(π₯) γ ππ₯=π(π₯) β«1βπ(π₯) ππ₯ββ«1β(π^β² (π₯) β«1βπ(π₯) ππ₯) ππ₯ Putting f(x) = log (log x) and g(x) = 1 I2=πππ(ππππ₯) β«1βγ1.ππ₯ββ«1β[π[πππ[ππππ₯]]/ππ₯ β«1βγ1.ππ₯γ] γ ππ₯ I2=log(ππππ₯)(π₯)ββ«1β[1/logβ‘π₯ .π(ππππ₯)/ππ₯ β«1βγ1.ππ₯γ]ππ₯ I2=π₯ πππ(ππππ₯)ββ«1βγ1/logβ‘π₯ .1/π₯ .π₯ ππ₯γ I2=π₯ πππ(ππππ₯)ββ«1βγ1/logβ‘π₯ .1/π₯ .π₯ ππ₯γ I2=π₯ πππ(ππππ₯)ββ«1βγ1/logβ‘π₯ ππ₯γ Solving ππ I3=β«1βγ1/logβ‘π₯ .ππ₯γ I3=β«1βγ1/logβ‘π₯ .1 ππ₯γ I3=1/ππππ₯ β«1βγ1.ππ₯ββ«1β[π(1/ππππ₯)/ππ₯ β«1βγ1.ππ₯γ] γ ππ₯ I3=1/ππππ₯.π₯ββ«1βγ(β1)/(ππππ₯)^2 (1/π₯).π₯ ππ₯γ Using by parts β«1βγπ(π₯) πβ‘(π₯) γ ππ₯=π(π₯) β«1βπ(π₯) ππ₯ββ«1β(π^β² (π₯) β«1βπ(π₯) ππ₯) ππ₯ Putting f(x) = 1/πππβ‘π₯ and g(x) = 1 I3=π₯/logβ‘π₯ +β«1βγ1/(logβ‘π₯ )^2 ππ₯γ Putting the value of I3 in I2 , we get I2=π₯ πππ(logβ‘π₯ )βπΌ3 I2=π₯ πππ(logβ‘π₯ )β[π₯/logβ‘π₯ +β«1β1/(logβ‘π₯ )^2 ππ₯] I2=π₯ πππ(logβ‘π₯ )βπ₯/logβ‘π₯ ββ«1βγ1/(logβ‘π₯ )^2 ππ₯γ Now, Putting the value of πΌ2 in πΌ1 , we get I1=I2+β«1βγ1/(logβ‘π₯ )^2 ππ₯γ I1=π₯ πππ(logβ‘π₯ )βπ₯/logβ‘π₯ ββ«1βγ1/(logβ‘π₯ )^2 ππ₯γ+β«1βγ1/(logβ‘π₯ )^2 ππ₯γ β΄ I1=π πππ(πππβ‘π )βπ/πππβ‘π + C