CBSE Class 12 Math 2020 Delhi Set 2 Solved Paper

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Question : 7 of 11
 
Marks: +1, -0
Find ∫x+1x(1−2x)dx
Solution:
Let
I=∫x+1x(1−2x)dx
Also, let
Also, let x+1x(1−2x)=Ax+B1−2x
⇒x+1x(1−2x)=A(1−2x)+Bxx(1−2x)
⇒x+1=x(−2A+B)+A
Comparing coefficient of x both sides, we get
1=−2A+B
Comparing constant term both sides, we get
A=1
∴ From equation (i),
1=−2(1)+B
⇒B=3
∴∫x+1x(1−2x)dx=∫1xdx+∫31−2xdx
=log|x|+3log|1−2x|−2+C
=log|x|−32log|1−2x|+C Ans.
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