A quadratic expression is defined in a variable $x$ and it is $x^2+6x+10$. The indefinite integration for the multiplicative inverse of this quadratic expression has to evaluate with respect to $x$ in this integral problem.

$\displaystyle \int{\dfrac{1}{x^2+6x+10}\,\,}dx$

In this integration problem, there is no linear expression in one variable in the numerator. Similarly, it is not possible to factorize (or factorise) the quadratic expression. Hence, it is recommendable to use completing the square method for finding the integration of the given rational expression.

Let’s try completing the square method to convert the quadratic expression in the denominator into square form completely.

$=\,\,\,$ $\displaystyle \int{\dfrac{1}{x^2+2 \times 3x+10}\,\,}dx$

The sign of the second term in the quadratic expression is positive. So, use the square of sum of two terms rule for transforming the quadratic expression into complete square form.

$=\,\,\,$ $\displaystyle \int{\dfrac{1}{x^2+2 \times 3 \times x+10}\,\,}dx$

$=\,\,\,$ $\displaystyle \int{\dfrac{1}{x^2+2 \times x \times 3+10}\,\,}dx$

In order to convert the quadratic expression into complete square form, add $3$ squared, to the quadratic expression and subtract the square of the number $3$ from the same polynomial. It helps us to transform the quadratic expression into square form completely.

$=\,\,\,$ $\displaystyle \int{\dfrac{1}{x^2+2 \times x \times 3+3^2-3^2+10}\,\,}dx$

$=\,\,\,$ $\displaystyle \int{\dfrac{1}{x^2+3^2+2 \times x \times 3-3^2+10}\,\,}dx$

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

Now, use the square of sum rule for converting the first three terms in square of a binomial.

$=\,\,\,$ $\displaystyle \int{\dfrac{1}{(x+3)^2-9+10}\,\,}dx$

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

For the finding the indefinite integration of the function, assume $y \,=\, x+3$ and differentiate this equation with respect to $x$.

$\implies$ $\dfrac{d}{dx}{(y)}$ $\,=\,$ $\dfrac{d}{dx}{(x+3)}$

$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\dfrac{d}{dx}{(x)}+\dfrac{d}{dx}{(3)}$

$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\dfrac{dx}{dx}+\dfrac{d}{dx}{(3)}$

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

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

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

$\implies$ $dy$ $\,=\,$ $dx$

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

Now, we can express the integral of the rational expression in $x$ into the integral of the rational expression in $y$.

$\implies$ $\displaystyle \int{\dfrac{1}{(x+3)^2+1}\,\,}dx$ $\,=\,$ $\displaystyle \int{\dfrac{1}{y^2+1}\,\,}dy$

$=\,\,\,$ $\displaystyle \int{\dfrac{1}{1+y^2}\,\,}dy$

The integration of the function can be evaluated by the integral rule of one by one plus square of a variable.

$=\,\,\,$ $\tan^{-1}{(y)}+c$

The solution for the integration of the function is evaluated in $y$ but the integration problem is given in $x$. Hence, convert the solution in $y$ in terms of $x$ by the appropriate replacement. We have assumed that $y \,=\, x+3$.

$=\,\,\,$ $\tan^{-1}{(x+3)}+c$

$\therefore \,\,\,$ $\displaystyle \int{\dfrac{1}{x^2+6x+10}\,\,}dx$ $\,=\,$ $\arctan{(x+3)}+c$

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