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  4. If ∫(dx/(x+2)(x2 +1)) = a log|1+x2| +b tan-1 x +(1/5)log|x+2|+C, then

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Q. If $ \int\frac{dx}{\left(x+2\right)\left(x^{2} +1\right)} = a log\left|1+x^{2}\right| +b tan^{-1} x +\frac{1}{5}log\left|x+2\right|+C$, then

Integrals

Solution:

We have, $I = \int\frac{dx}{\left(x+2\right)\left(x^{2}+1\right)}$

Let $ \frac{1}{\left(x+2\right)\left(x^{2}+1\right)} = \frac{A}{x+2} + \frac{Bx+C}{x^{2}+1}$

$ \Rightarrow 1= A\left(x^{2} +1\right) + Bx\left(x+2\right) + C\left(x+2\right) \quad...\left(i\right)$

Put $x = 0$ in $\left(i\right)$, we get $A + 2C = 1$

Put $x= -2$ in $\left(i\right)$, we get $A= \frac{1}{5}$

$\Rightarrow C =\frac{2}{5}$

Put $x= 1$ in $\left(i\right)$, we get

$ 1 = 2A + 3B + 3 C$, we get $B = \frac{-1}{5}$

$ \Rightarrow \int \frac{dx}{\left(x+2\right)\left(x^{2}+1\right)} = \frac{1}{5} \int \frac{dx}{x+2} -\frac{1}{5} \int \frac{xdx}{x^{2}+1} + \frac{2}{5} \int \frac{dx}{x^{2}+1}$

$ = \frac{1}{5 } log\left|x+2\right| -\frac{1}{10 }log\left|x^{2}+1\right| +\frac{2}{5 }tan^{-1} x + C $

Hence, $a= \frac{-1}{10}$ and $b = \frac{2}{5}$