Mathematics

$$ \int $$ $$\dfrac {dx}{(x^{2}+1)(x+3)}$$


SOLUTION
We have,
$$ I=\int \dfrac {1}{(x^{2}+1)(x+3)}$$

Let
$$\dfrac {1}{(x^{2}+1)(x+3)}=\dfrac{Ax+B}{x^2+1}+\dfrac{C}{x+3}$$

$$1=(Ax+B)(x+3)+C(x^2+1)$$

$$1=Ax^2+Bx+3Ax+3B+Cx^2+C$$             $$............(1)$$

On solving equation $$(1)$$, we get
$$A=-\dfrac{1}{10}, B=\dfrac{3}{10}, C=\dfrac{1}{10}$$

Therefore,
$$I=\displaystyle \int \dfrac{1}{10}\left(\dfrac{-x+3}{x^2+1}\right)+\dfrac{1}{10(x+3)}\ dx$$

$$I=- \dfrac{1}{10}\int \left(\dfrac{x}{x^2+1}\right)dx+\dfrac{3}{10}\int \dfrac{dx}{x^2+1}+\int \dfrac{1}{10(x+3)}\ dx$$

$$I=- \dfrac{1}{20}\int \left(\dfrac{2x}{x^2+1}\right)dx+\dfrac{3}{10}\int \dfrac{dx}{x^2+1}+\int \dfrac{1}{10(x+3)}\ dx$$

$$I=- \dfrac{1}{20}\ln(x^2+1)+\dfrac{3}{10}\tan^{-1}(x^2+1)+ \dfrac{1}{10}\ln (x+3)+C$$

Hence, this is the answer.
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