CBSE Class 12 Sample Paper for 2022 Boards (For Term 2)
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CBSE Class 12 Sample Paper for 2022 Boards (For Term 2)
Last updated at Dec. 14, 2024 by Teachoo
This question is similar to Ex 7.2, 35 - Chapter 7 Class 12 - Integrals
Transcript
Question 1 β Choice 1 Find β«1βγlogβ‘π₯/(1 + logβ‘π₯ )^2 ππ₯γLet π=β«1βγlogβ‘π₯/(1 + logβ‘π₯ )^2 ππ₯γ =β«1βγ(πππβ‘π + π β π)/(1 + logβ‘π₯ )^2 ππ₯γ =β«1βγ((1 + logβ‘π₯) β 1)/(1 + logβ‘π₯ )^2 ππ₯γ =β«1βγ((1 + logβ‘π₯) )/(1 + logβ‘π₯ )^2 ππ₯γββ«1βγ1/(1 + logβ‘π₯ )^2 ππ₯γ =β«1βγ(π )/((π + πππβ‘π ) ) π πγββ«1βγπ/(π + πππβ‘π )^π π πγ Solving β«1βγ(π )/((π + π₯π¨π β‘π± ) ) ππ±γ Using Integration by parts β«1βγ(π )/((π + πππβ‘π ) ) π πγ = β«1βγ(1 )/((1 + πππβ‘π₯ ) ) Γ 1 ππ₯γ = 1/((1 + logβ‘π₯)) β«1βγ1 ππ₯γββ«1β(π (π/(π + πππ π))/π π β«1βγπ π πγ) π π = 1/((1 + logβ‘π₯))Γ π₯ββ«1β((βπ)/(π +πππ π)^π Γπ/π Γ π) π π = π₯/((1 + logβ‘π₯))+β«1βπ/(π + πππ π)^π π πWe know that 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/(1 + log x) and g(x) = 1 Thus I=β«1βγ(1 )/((1 + πππβ‘π₯ ) ) ππ₯γββ«1βγ1/(1 + πππβ‘π₯ )^2 ππ₯γ = π₯/((1 + logβ‘π₯))+β«1β1/(1 + πππ π₯)^2 ππ₯ββ«1βγ1/(1 + πππβ‘π₯ )^2 ππ₯γ = π/((π + πππβ‘π))+πͺ
