CBSE Class 12 Sample Paper for 2022 Boards (For Term 2)

Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

## Find ∫1▒γ(x + 1)/((x^2  + 1)  x) dxγ

This question is similar to Ex 13.2, 2 - Chapter 13 Class 12 - Probability

### Transcript

Question 7 Find β«1βγ(π₯ + 1)/((π₯^2 + 1) π₯) ππ₯γLet (π₯ + 1)/((π₯^2 + 1) π₯) = (π¨π + π©)/((π^π + π) ) + πͺ/π (π₯ + 1)/((π₯^2 + 1) π₯) = ((π΄π₯ + π΅)π₯ + πΆ(1 + π₯^2 ))/((π₯^2 + 1) π₯) By cancelling denominator π + π = (π¨π + π©)π + πͺ(π + π^π ) Putting π=π 0 + 1 = (π΄(0) + π΅) Γ 0 + πΆ(1 +0^2 ) 1 = πΆ πͺ = π Putting π=π 1 + 1 = (π΄(1) + π΅) Γ 1 + πΆ(1 +1^2 ) 2 = (π΄ + π΅) +2πΆ Putting πΆ = 1 2 = (π΄ + π΅) +2 Γ 1 2 = (π΄ + π΅) +2 2β2 = (π΄ + π΅) 0 = π΄ + π΅ π¨=β π© Putting π=βπ β1 + 1 = (π΄(β1) + π΅) Γ β1 + πΆ(1 +γ(β1)γ^2 ) 0 = β(βπ΄ + π΅) +πΆ Γ (1+1) 0 = β(βπ΄ + π΅) +2πΆ Putting π΄=β π΅ 0 = β(π΅ + π΅) +2πΆ 0 = β2B +2πΆ 2B =2πΆ B =πΆ Putting πΆ = 1 π© = π And, π΄=βπ΅ β΄ π¨=βπ Thus, π΄=β1, π΅=1, πΆ = 1 So, we can write (π + π)/((π^π + π) π) = (π΄π₯ + π΅)/((π₯^2 + 1) ) + πΆ/π₯ = ((β1)π₯ +1)/((π₯^2 + 1) ) + 1/π₯ = (βπ + π)/((π^π + π) ) + π/π Therefore integrating β«1β(π₯ + 1)/((π₯^2 + 1) π₯) ππ₯ = β«1β(βπ + π)/((π^π + π) ) ππ₯ + β«1β1/(π₯ ) ππ₯ = β«1β(βπ₯ + 1)/((π₯^2 + 1) ) ππ₯ + β«1β1/(π₯ ) ππ₯ = β«1β(βπ₯)/((π₯^2 + 1) ) ππ₯ + β«1β1/(π₯^2 + 1) ππ₯ + β«1β1/(π₯ ) ππ₯ = β«1β(βπ₯)/((π₯^2 + 1) ) ππ₯ + γπ­ππ§γ^(βπ)β‘π + πππβ‘γ|π|γ+πΆ = β«1β(βπ₯)/((π₯^2 + 1) ) ππ₯ + γπ­ππ§γ^(βπ)β‘π + πππβ‘γ|π|γ+πΆ Solving π1 I1 = β«1β(βπ₯)/(π₯^2 + 1) ππ₯ Let π = π^π+π ππ‘/ππ₯ = 2π₯ ππ‘/2π₯ = ππ₯ Hence β«1β(βπ₯)/(π₯^2 + 1) ππ₯ = β«1βγ(βπ₯)/π‘ . ππ‘/2π₯γ = ββ«1βππ‘/2(π‘) = (β1)/2 γlog γβ‘|π‘|+πΆ1 Putting back t = π₯^2+1 = (βπ)/π γπππ γβ‘|π^π+π|+πͺπ Therefore integrating β«1β(π₯ + 1)/((π₯^2 + 1) π₯) ππ₯ = (βπ)/π γπππ γβ‘|π^π+π| + γπ­ππ§γ^(βπ)β‘π + πππβ‘γ|π|γ+πΆ