The linear expression in one variable $2x+3$ is divided by a quadratic expression $x^2+4x+5$ to express a quantity in rational form.

$\displaystyle \int{\dfrac{2x+3}{x^2+4x+5}\,}dx$

The indefinite integral of this rational function has to calculate with respect to $x$ in this problem. Now, let’s learn how to find the integral of the given rational function in calculus.

The numerator $2x+3$ is a linear expression in one variable and the denominator $x^2+4x+5$ is a quadratic expression. The numerator should be adjusted as the derivative of the denominator in order to find the integral of this type of rational functions.

The derivative of quadratic expression is $2x+4$ but the numerator $2x+3$ is slightly different to the derivative. Hence, add $1$ to the expression in the numerator and subtract same quantity from their sum.

$=\,\,\,$ $\displaystyle \int{\dfrac{2x+3+1-1}{x^2+4x+5}\,}dx$

$=\,\,\,$ $\displaystyle \int{\dfrac{2x+4-1}{x^2+4x+5}\,}dx$

The rational function can be split as the difference of two rational functions as follows.

$=\,\,\,$ $\displaystyle \int{\bigg(\dfrac{2x+4}{x^2+4x+5}-\dfrac{1}{x^2+4x+5}\bigg)\,}dx$

According to the difference rule of integration, the integral of difference of functions is equal to the difference of their integrals.

$=\,\,\,$ $\displaystyle \int{\dfrac{2x+4}{x^2+4x+5}\,}dx$ $-$ $\displaystyle \int{\dfrac{1}{x^2+4x+5}\,}dx$

In this step, let’s focus on finding the integral of the rational function in the first term of the expression.

$\displaystyle \int{\dfrac{2x+4}{x^2+4x+5}\,}dx$ $-$ $\displaystyle \int{\dfrac{1}{x^2+4x+5}\,}dx$

$=\,\,\,$ $\displaystyle \int{\dfrac{(2x+4) \times dx}{x^2+4x+5}\,}$ $-$ $\displaystyle \int{\dfrac{1}{x^2+4x+5}\,}dx$

Suppose $u \,=\, x^2+4x+5$ and differentiate both sides with respect to $x$.

$\implies$ $\dfrac{d}{dx}{(u)}$ $\,=\,$ $\dfrac{d}{dx}{(x^2+4x+5)}$

As per the addition rule of derivatives, the derivative of sum of the terms is equal to sum of their derivatives.

$\implies$ $\dfrac{du}{dx}$ $\,=\,$ $\dfrac{d}{dx}{(x^2)}$ $+$ $\dfrac{d}{dx}{(4x)}$ $+$ $\dfrac{d}{dx}{(5)}$

Find the derivative of $x$ square as per the power rule of derivatives and also find the derivative of number $5$ as per the derivative rule of a constant.

$\implies$ $\dfrac{du}{dx}$ $\,=\,$ $2x^{2-1}$ $+$ $\dfrac{d}{dx}{(4 \times x)}$ $+$ $0$

Use the constant multiple rule of derivatives to separate the constant factor $4$ from the differentiation.

$\implies$ $\dfrac{du}{dx}$ $\,=\,$ $2x^{1}$ $+$ $4 \times \dfrac{d}{dx}{(x)}$

Finally, find the derivative of $x$ with respect to $x$ as per the derivative rule of a variable.

$\implies$ $\dfrac{du}{dx}$ $\,=\,$ $2x$ $+$ $4 \times 1$

$\implies$ $\dfrac{du}{dx}$ $\,=\,$ $2x+4$

$\,\,\,\therefore\,\,\,\,\,\,$ $du$ $\,=\,$ $(2x+4) \times dx$

We have assumed that $u \,=\, x^2+4x+5$ and also derived that $du$ $\,=\,$ $(2x+4) \times dx$. So, substitute them in the first term of the expression to convert the expression in the first term in terms of $u$.

$\implies$ $\displaystyle \int{\dfrac{(2x+4) \times dx}{x^2+4x+5}\,}$ $-$ $\displaystyle \int{\dfrac{1}{x^2+4x+5}\,}dx$ $\,=\,$ $\displaystyle \int{\dfrac{du}{u}\,}$ $-$ $\displaystyle \int{\dfrac{1}{x^2+4x+5}\,}dx$

$=\,\,\,$ $\displaystyle \int{\dfrac{1 \times du}{u}\,}$ $-$ $\displaystyle \int{\dfrac{1}{x^2+4x+5}\,}dx$

$=\,\,\,$ $\displaystyle \int{\dfrac{1}{u}\, \times du}$ $-$ $\displaystyle \int{\dfrac{1}{x^2+4x+5}\,}dx$

$=\,\,\,$ $\displaystyle \int{\dfrac{1}{u}\, du}$ $-$ $\displaystyle \int{\dfrac{1}{x^2+4x+5}\,}dx$

Use the reciprocal rule of integrals to find the integral of the multiplicative inverse of $u$ with respect to $u$.

$=\,\,\,$ $\log_{e}{|u|}+c_1$ $-$ $\displaystyle \int{\dfrac{1}{x^2+4x+5}\,}dx$

Now, substitute the value of $u$ to get the first term in terms of $x$.

$=\,\,\,$ $\log_{e}{\big|x^2+4x+5\big|}+c_1$ $-$ $\displaystyle \int{\dfrac{1}{x^2+4x+5}\,}dx$

It is time to calculate the indefinite integral of the remaining rational function. The rational function is a reciprocal of quadratic expression. It can be evaluated by completing the square method.

$=\,\,\,$ $\log_{e}{\big|x^2+4x+5\big|}+c_1$ $-$ $\displaystyle \int{\dfrac{1}{x^2+2 \times 2 \times x+5}\,}dx$

$=\,\,\,$ $\log_{e}{\big|x^2+4x+5\big|}+c_1$ $-$ $\displaystyle \int{\dfrac{1}{x^2+2 \times 2 \times x+4+1}\,}dx$

$=\,\,\,$ $\log_{e}{\big|x^2+4x+5\big|}+c_1$ $-$ $\displaystyle \int{\dfrac{1}{x^2+2 \times 2 \times x+2^2+1}\,}dx$

$=\,\,\,$ $\log_{e}{\big|x^2+4x+5\big|}+c_1$ $-$ $\displaystyle \int{\dfrac{1}{(x+2)^2+1}\,}dx$

$=\,\,\,$ $\log_{e}{\big|x^2+4x+5\big|}+c_1$ $-$ $\displaystyle \int{\dfrac{1}{1+(x+2)^2}\,}dx$

Suppose $y \,=\, x+2$. Now, differentiate the equation with respect to $x$.

$\implies$ $\dfrac{d}{dx}{(y)}$ $\,=\,$ $\dfrac{d}{dx}{(x+2)}$

$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\dfrac{d}{dx}{(x)}$ $+$ $\dfrac{d}{dx}{(2)}$

$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $1$ $+$ $0$

$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $1$

$\implies$ $dy$ $\,=\,$ $1 \times dx$

$\,\,\,\therefore\,\,\,\,\,\,$ $dx$ $\,=\,$ $dy$

In this step, we took $y \,=\, x+2$ and derived that $dx \,=\, dy$. Now, transform the integral function in terms of $y$ by substituting these values.

$=\,\,\,$ $\log_{e}{\big|x^2+4x+5\big|}+c_1$ $-$ $\displaystyle \int{\dfrac{1}{1+y^2}\,}dy$

According to the reciprocal integral rule in one plus square, the integral of the reciprocal of $1$ plus square of $y$ with respect to $y$ is equal to inverse tangent of $y$.

$=\,\,\,$ $\log_{e}{\big|x^2+4x+5\big|}+c_1$ $-$ $\big(\arctan{(y)}+c_2\big)$

It can also be written in the following from as per the inverse trigonometry.

$=\,\,\,$ $\log_{e}{\big|x^2+4x+5\big|}+c_1$ $-$ $\big(\tan^{-1}{(y)}+c_2\big)$

$=\,\,\,$ $\log_{e}{\big|x^2+4x+5\big|}+c_1$ $-$ $\tan^{-1}{(y)}$ $-$ $c_2$

$=\,\,\,$ $\log_{e}{\big|x^2+4x+5\big|}$ $-$ $\tan^{-1}{(y)}$ $+$ $c_1$ $-$ $c_2$

Now, substitute the value of $y$ to get the integral of the given rational function in terms of $x$.

$=\,\,\,$ $\log_{e}{\big|x^2+4x+5\big|}$ $-$ $\tan^{-1}{(x+2)}$ $+$ $c_1$ $-$ $c_2$

The difference of the constants can be easily denoted by a constant $c$.

$=\,\,\,$ $\log_{e}{\big|x^2+4x+5\big|}$ $-$ $\tan^{-1}{(x+2)}$ $+$ $c$

It is also written in another form in integral calculus as follows.

$=\,\,\,$ $\log_{e}{\big|x^2+4x+5\big|}$ $-$ $\arctan{(x+2)}$ $+$ $c$

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