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  • If \int \frac{d x}{(x+2)\left(x^{2}+1\right)}=a \log \left|1+x^{2}\right|+b \tan ^{-1} x+\frac{1}{5} \log |x+2|+c log |x + 2| + C, then A. a equals minus1/10

If \int \frac{d x}{(x+2)\left(x^{2}+1\right)}=a \log \left|1+x^{2}\right|+b \tan ^{-1} x+\frac{1}{5} \log |x+2|+c then

\\ A.a=\frac{-1}{10}, b=\frac{-2}{5}\\ B. a=\frac{1}{10}, b=-\frac{2}{5}\\ C. \mathrm{a}=\frac{-1}{10}, \mathrm{~b}=\frac{2}{5}\\ D. a=\frac{1}{10}, b=\frac{2}{5}\\

Answers (1)

C)

Given:\int \frac{d x}{(x+2)\left(x^{2}+1\right)}=a \log \left|1+x^{2}\right|+b \tan ^{-1} x+\frac{1}{5} \log |x+2|+c

Using concept of partial fractions

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

\\ \Rightarrow x^{2}(A+B)+x(C+2 B)+(A+2 C)=1 \\ \Rightarrow A+B=0 \quad \ldots(1) \\ \Rightarrow C+2 B=0 \quad \ldots(2) \\ \Rightarrow A+2 C=1...(3)

On solving the above three equations we get

\\ \Rightarrow A=\frac{1}{5}, B=-\frac{1}{5} \text { and } C=\frac{2}{5} \\ \Rightarrow \frac{1}{(x+2)\left(x^{2}+1\right)}=\frac{\frac{1}{5}}{(x+2)}+\frac{-\frac{1}{5} x+\frac{2}{5}}{\left(x^{2}+1\right)} \\ \Rightarrow \frac{\frac{1}{5}}{(x+2)}+\frac{-\frac{1}{5} x+\frac{2}{5}}{\left(x^{2}+1\right)}=\frac{1}{5(x+2)}-\frac{x}{5\left(x^{2}+1\right)}+\frac{2}{5\left(x^{2}+1\right)}

\\ \int \frac{d x}{(x+2)\left(x^{2}+1\right)}=\int\left(\frac{1}{5(x+2)}-\frac{x}{5\left(x^{2}+1\right)}+\frac{2}{5\left(x^{2}+1\right)}\right) d x \\ =\frac{1}{5} \ln |x+2|-\frac{1}{10} \ln \left|x^{2}+1\right|+\frac{2}{5} \tan ^{-1} x +c \ldots (2)

On comparing (1) and (2) we get,

\Rightarrow a=-\frac{1}{10} \text { and } b=\frac{2}{5}

Posted by

infoexpert22

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