Test Index

CBSE Class 12 Math 2020 Delhi Set 2 Solved Paper

© examsnet.com
Question : 7 of 11
Marks: +1, -0
Find ∫  x+1x(1−2x)dx\int\;\frac{x+1}{x(1-2x)} dx
Solution:  
Let
I=∫  x+1x(1−2x)dxI=\int\;\frac{x+1}{x(1-2x)} dx
Also, let
    Also, let       x+1x(1−2x)=Ax+B1−2x\;\;\text{Also, let }\;\;\;\frac{x+1}{x(1-2x)}=\frac{A}{x}+\frac{B}{1-2x}
⇒  x+1x(1−2x)=A(1−2x)+Bxx(1−2x)\Rightarrow\;\frac{x+1}{x(1-2x)}=\frac{A(1-2x)+Bx}{x(1-2x)}
⇒  x+1=x(−2A+B)+A\Rightarrow\;x+1=x(-2A+B)+A
Comparing coefficient of xx both sides, we get
1=−2A+B1=-2A+B
Comparing constant term both sides, we get
A=1A=1
∴    \therefore\;\; From equation (i),
1=−2(1)+B1=-2(1)+B
⇒    B=3\Rightarrow\;\;B=3
∴    ∫  x+1x(1−2x)dx=∫  1xdx+∫  31−2xdx\therefore\;\;\int\;\frac{x+1}{x(1-2x)}dx=\int\;\frac{1}{x}dx+\int\;\frac{3}{1-2x}dx
=log⁡∣x∣+3log⁡∣1−2x∣−2+C=\log|x|+\frac{3\log|1-2x|}{-2}+C
=log⁡∣x∣−32log⁡∣1−2x∣+C  Ans.  =\log|x|-\frac{3}{2}\log|1-2x|+C\;\text{Ans.}\;
$ $
© examsnet.com
Go to Question:
Prev Question
Next Question