# Evaluate $\displaystyle \int{\dfrac{1}{x^2+4x+3} \,} dx$

The indefinite integral of a rational expression has to evaluate with respect to $x$ in this problem, where the rational expression is a reciprocal of a quadratic expression $x^2+4x+3$

$\displaystyle \int{\dfrac{1}{x^2+4x+3} \,} dx$

The algebraic expression in the denominator of the rational expression is a quadratic expression. In fact, it is not possible to evaluate the indefinite integration of this algebraic expression directly. Hence, we have to factorize the quadratic expression firstly and it will be useful to us in splitting the rational expression as the partial fractions. The quadratic expression $x^2+4x+3$ can be factored by splitting its middle term.

$= \,\,\,$ $\displaystyle \int{\dfrac{1}{x^2+3x+x+3} \,} dx$

$= \,\,\,$ $\displaystyle \int{\dfrac{1}{x(x+3)+1(x+3)} \,} dx$

$= \,\,\,$ $\displaystyle \int{\dfrac{1}{(x+3)(x+1)} \,} dx$

$= \,\,\,$ $\displaystyle \int{\dfrac{1}{(x+1)(x+3)} \,} dx$

### Split function by Partial fractions Decomposition

The function is a rational expression that consists of non-repeated linear factors in denominator. According to partial fractions decomposition, it can be decomposed into two partial fractions.

$\implies$ $\dfrac{1}{(x+1)(x+3)}$ $\,=\,$ $\dfrac{A}{x+1}+\dfrac{B}{x+3}$

$(1)\,\,\,$ Let $x \,=\, -1$, then $A$ $\,=\,$ $\dfrac{1}{-1+3}$ $\,=\,$ $\dfrac{1}{2}$

$(2)\,\,\,$ Let $x \,=\, -3$, then $B$ $\,=\,$ $\dfrac{1}{-3+1}$ $\,=\,$ $\dfrac{1}{-2}$ $\,=\,$ $-\dfrac{1}{2}$

Now, substitute the values of constants $A$ and $B$ and then simplify the expression.

$\implies$ $\dfrac{1}{(x+1)(x+3)}$ $\,=\,$ $\dfrac{\Big(\dfrac{1}{2}\Big)}{x+1}+\dfrac{\Big(-\dfrac{1}{2}\Big)}{x+3}$

$\implies$ $\dfrac{1}{(x+1)(x+3)}$ $\,=\,$ $\dfrac{1 \times \Big(\dfrac{1}{2}\Big)}{x+1}+\dfrac{1 \times \Big(-\dfrac{1}{2}\Big)}{x+3}$

$\implies$ $\dfrac{1}{(x+1)(x+3)}$ $\,=\,$ $\Big(\dfrac{1}{2}\Big) \times \dfrac{1}{x+1}+\Big(-\dfrac{1}{2}\Big) \times \dfrac{1}{x+3}$

$\implies$ $\dfrac{1}{(x+1)(x+3)}$ $\,=\,$ $\dfrac{1}{2} \times \dfrac{1}{x+1}-\dfrac{1}{2} \times \dfrac{1}{x+3}$

### Evaluate Indefinite Integration of each function

Now, it is time to concentrate on evaluating the indefinite integral of the given rational expression.

$\implies$ $\displaystyle \int{\dfrac{1}{(x+1)(x+3)}\,}dx$ $\,=\,$ $\displaystyle \int{\Bigg(\dfrac{1}{2} \times \dfrac{1}{x+1}-\dfrac{1}{2} \times \dfrac{1}{x+3}\Bigg)\,}dx$

As per the difference rule of integration, it can be simplified further.

$=\,\,\,$ $\displaystyle \int{\Bigg(\dfrac{1}{2} \times \dfrac{1}{x+1}\Bigg)\,}dx$ $-$ $\displaystyle \int{\Bigg(\dfrac{1}{2} \times \dfrac{1}{x+3}\Bigg)\,}dx$

The constant can be separated from the integral function by the constant multiple rule of integration.

$=\,\,\,$ $\displaystyle \dfrac{1}{2} \times \int{\dfrac{1}{x+1}\,}dx$ $-$ $\displaystyle \dfrac{1}{2} \times \int{\dfrac{1}{x+3}\,}dx$

Each term can be integrated by the integral rule of reciprocal of linear expression in one variable.

$=\,\,\,$ $\dfrac{1}{2} \times \dfrac{1}{1} \times \Big(\log_e{|x+1|}+c_1\Big)$ $-$ $\dfrac{1}{2} \times \dfrac{1}{1} \times \Big(\log_e{|x+3|}+c_2\Big)$

$=\,\,\,$ $\dfrac{1}{2} \times 1 \times \Big(\log_e{|x+1|}+c_1\Big)$ $-$ $\dfrac{1}{2} \times 1 \times \Big(\log_e{|x+3|}+c_2\Big)$

$=\,\,\,$ $\dfrac{1}{2} \times \Big(\log_e{|x+1|}+c_1\Big)$ $-$ $\dfrac{1}{2} \times \Big(\log_e{|x+3|}+c_2\Big)$

$=\,\,\,$ $\dfrac{1}{2} \times \log_e{|x+1|} + \dfrac{1}{2} \times c_1$ $-$ $\dfrac{1}{2} \times \log_e{|x+3|}- \dfrac{1}{2} \times c_2$

$=\,\,\,$ $\dfrac{1}{2} \times \log_e{|x+1|} + \dfrac{1 \times c_1}{2}$ $-$ $\dfrac{1}{2} \times \log_e{|x+3|}- \dfrac{1 \times c_2}{2}$

$=\,\,\,$ $\dfrac{1}{2} \times \log_e{|x+1|} + \dfrac{c_1}{2}$ $-$ $\dfrac{1}{2} \times \log_e{|x+3|}-\dfrac{c_2}{2}$

$=\,\,\,$ $\dfrac{1}{2} \times \log_e{|x+1|}$ $-$ $\dfrac{1}{2} \times \log_e{|x+3|}$ $+$ $\dfrac{c_1}{2}$ $-$ $\dfrac{c_2}{2}$

$=\,\,\,$ $\dfrac{1}{2} \times \Big(\log_e{|x+1|}-\log_e{|x+3|}\Big)$ $+$ $\dfrac{c_1-c_2}{2}$

The difference of the logarithmic terms can be simplified by the quotient rule of logarithms. The second term is a constant and it can be simplify denoted by $c$.

$=\,\,\,$ $\dfrac{1}{2} \times \log_e{\Bigg|\dfrac{x+1}{x+3}\Bigg|}+c$

$=\,\,\,$ $\dfrac{1}{2}\log_e{\Bigg|\dfrac{x+1}{x+3}\Bigg|}+c$

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