Math Doubts

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

A rational expression is formed by the quotient of two algebraic expressions $x^2+1$ and $x^4+1$. The indefinite integral of this rational expression has to evaluate with respect to $x$ in this integration problem.

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

Simplify the Rational algebraic expression

The degree of the polynomial $x^2+1$ is $2$ and the degree of the polynomial $x^4+1$ is $4$. So, it is not possible to divide the quadratic expression by the quartic expression. Hence, we must focus on simplifying each expression in both numerator and denominator of the rational expression.

Mathematically, the quadratic expression $x^2+1$ cannot be simplified but the bi-quadratic expression $x^4+1$ with an acceptable mathematical adjustment can be simplified. The term $x^4$ can be split into two factors by the product rule of exponents.

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

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

The two algebraic expressions are purely expressed in square expression form. For simplifying each algebraic expression, express $1$ in same form.

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

The rational expression in algebraic form can be simplified further by taking the common factor $x^2$ out from the expressions.

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

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

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

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

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

Converting the expression in Square form

In the denominator, the expression contains two terms in square form and the whole expression can be converted into square form but the number $2$ must be included in the expression acceptably.

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

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

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

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

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

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

Now, the algebraic expression in the denominator can be simplified by the square of difference rule.

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

Evaluate Integration of the Rational expression

The rational expression is simplified to maximum level and it cannot be done further. So, we have to focus on the integration of the function. In fact, the rational expression is not in the form of any integral rule. However, the expression in the numerator can be obtained by differentiating the expression $x-\dfrac{1}{x}$.

Take, $u \,=\, x-\dfrac{1}{x}$ and then differentiate it both sides with respect to $x$.

$\implies$ $\dfrac{d}{dx}{\,(u)}$ $\,=\,$ $\dfrac{d}{dx}{\,\Big(x-\dfrac{1}{x}\Big)}$

$\implies$ $\dfrac{du}{dx}$ $\,=\,$ $\dfrac{d}{dx}{\,(x)}$ $-$ $\dfrac{d}{dx}{\,\Big(\dfrac{1}{x}\Big)}$

$\implies$ $\dfrac{du}{dx}$ $\,=\,$ $1-\Big(-\dfrac{1}{x^2}\Big)$

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

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

Now, eliminate the $x$ by replacing it with its equivalent value in $u$ for performing the indefinite integration.

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

Now, let’s evaluate the indefinite integration of the rational expression in $u$.

$=\,\,\,$ $\displaystyle \int{\dfrac{du}{u^2+2}\,}$

The number $2$ in denominator can be written in square form for using the integration rule for reciprocal of sum of squares.

$=\,\,\,$ $\displaystyle \int{\dfrac{du}{u^2+(\sqrt{2)}^2}\,}$

Finally, use the reciprocal integration rule of sum of squares formula for evaluating the indefinite integration of reciprocal of sum of squares.

$=\,\,\,$ $\dfrac{1}{\sqrt{2}}\tan^{-1}{\Big(\dfrac{u}{\sqrt{2}}\Big)}+c$

The given function in the integral problem is in $x$. So, express the solution of the integral function in $x$ by replacing the value of $u$ in $x$.

$=\,\,\,$ $\dfrac{1}{\sqrt{2}}\tan^{-1}{\Bigg(\dfrac{x-\dfrac{1}{x}}{\sqrt{2}}\Bigg)}+c$

$=\,\,\,$ $\dfrac{1}{\sqrt{2}}\tan^{-1}{\Bigg(\dfrac{x^2-1}{\sqrt{2}x}\Bigg)}+c$

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