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Evaluate the integral of  \int\frac{x+2}{x^{2}\left ( x+1 \right )}dx

 

Option: 1

-ln\left | x \right |-\frac{2}{x}-ln\left | x+1 \right |+C


Option: 2

ln\left | x \right |+\frac{2}{x}+ln\left | x+1 \right |+C


Option: 3

-ln\left | x \right |+\frac{2}{x}+ln\left | x+1 \right |+C


Option: 4

-ln\left | x \right |-\frac{2}{x}+ln\left | x+1 \right |+C


Answers (1)

Given integral,

\int\frac{x+2}{x^{2}\left ( x+1 \right )}dx

\int\frac{x+2}{x^{2}\left ( x+1 \right )}dx=\int\left (\frac{A}{x}+\frac{B}{x^{2}} +\frac{C}{x+1}\right)dx

Using partial fractions,

Ax\left ( x+1 \right )+B\left ( \left ( x+1 \right ) \right )+Cx^{2}=x+2

Put x=0,

B=2

Put x=-1,

c=1

x=1,

2A+2B+C=3\Rightarrow A=-1

Now,

\int\frac{x+2}{x^{2}\left ( x+1 \right )}dx=\int\left (\frac{-1}{x}+\frac{2}{x^{2}} +\frac{1}{x+1}\right)dx

\int\frac{x+2}{x^{2}\left ( x+1 \right )}dx=-ln\left | x \right |+\frac{2x^{-1}}{\left ( -1 \right )}+ln\left | x+1 \right |+C

\int\frac{x+2}{x^{2}\left ( x+1 \right )}dx=-ln\left | x \right |-\frac{2}{x}+ln\left | x+1 \right |+C

Posted by

Ramraj Saini

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