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

In this problem, a quadratic expression is divided by a quartic expression. The indefinite integral of the rational expression has to evaluate with respect to $x$.

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

In numerator, the number $1$ is subtracted from the square of a variable $x$ and it is added to the $x$ raised to the power $4$ in the denominator. Let’s learn how to find the indefinite integration of the quotient of $x^2-1$ by $x^4+1$ with respect to $x$ in this indefinite integration problem.

### Cancel the square term from expressions

There is square term in the expression of the numerator and there is a quartic expression in the denominator. So, the quartic term can be written as a product of two square factors. This idea helps us to cancel the square factors from both expressions. Hence, split the quartic term as a product of two square factors by using the product rule of exponents.

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

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

For cancelling the square factors of both expressions, write the constant term in both expressions as a quotient of same square terms. This trick clears the route for cancelling the square factors.

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

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

Now, each term in both expressions consist of a square factor. So, the factor in square form can be taken out common from each expression.

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

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

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

### Adjust the denominator as a square of a Binomial

Two square terms in sum form forms an expression in the denominator. It can be converted into square of a sum basis binominal but it should contain a number $2$ in sum form. Hence, add $2$ to expression and subtract the same quantity from the expression for maintaining the mathematical balance.

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

It is time to concentrate on converting the mathematical expression as a square of a binomial as follows.

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

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

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

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

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

### Find the Indefinite integration of the function

The expression in the numerator can be obtaining by differentiating the binomial in square form in the denominator.

Take $l \,=\, x+\dfrac{1}{x}$

Now, differentiate the algebraic equation with respect to $x$.

$\implies$ $\dfrac{d}{dx}{(l)}$ $\,=\,$ $\dfrac{d}{dx}{\bigg(x+\dfrac{1}{x}\bigg)}$

$\implies$ $\dfrac{dl}{dx}$ $\,=\,$ $\dfrac{d}{dx}{\bigg(x+\dfrac{1}{x}\bigg)}$

$\implies$ $\dfrac{dl}{dx}$ $\,=\,$ $\dfrac{d}{dx}{(x)}$ $+$ $\dfrac{d}{dx}{\bigg(\dfrac{1}{x}\bigg)}$

The derivative of a variable is one as per derivative rule of a variable. The derivative of a reciprocal of a variable can also be evaluated by the reciprocal rule of the differentiation.

$\implies$ $\dfrac{dl}{dx}$ $\,=\,$ $1$ $-$ $\dfrac{1}{x^2}$

$\,\,\,\therefore\,\,\,\,\,\,$ $dl$ $\,=\,$ $\bigg(1-\dfrac{1}{x^2}\bigg) \times dx$

Now, transform the integral of the function in terms of $x$ into the integral of the function in terms of $l$.

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

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

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

Now, let’s find the indefinite integration of the function.

$\displaystyle \int{\dfrac{1}{l^2-2}}\,dl$

The integral of the function is similar to the integral reciprocal rule of the difference of squares.

$=\,\,\,$ $\displaystyle \int{\dfrac{1}{l^2-\big(\sqrt{2}\big)^2}}\,dl$

$=\,\,\,$ $\dfrac{1}{2\sqrt{2}}\log_e{\Bigg|\dfrac{l-\sqrt{2}}{l+\sqrt{2}}\Bigg|}+c$

Now, substitute the value of $l$ in terms of $x$ to express the solutions of the indefinite integral problem.

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

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

$=\,\,\,$ $\dfrac{1}{2\sqrt{2}}\log_e{\Bigg|\dfrac{\dfrac{x^2+1-\sqrt{2}x}{x}}{\dfrac{x^2+1+\sqrt{2}x}{x}}\Bigg|}+c$

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

$=\,\,\,$ $\dfrac{1}{2\sqrt{2}}\log_e{\Bigg|\dfrac{x^2+1-\sqrt{2}x}{\cancel{x}} \times \dfrac{\cancel{x}}{x^2+1+\sqrt{2}x}\Bigg|}+c$

$=\,\,\,$ $\dfrac{1}{2\sqrt{2}}\log_e{\Bigg|\dfrac{x^2+1-\sqrt{2}x}{x^2+1+\sqrt{2}x}\Bigg|}+c$

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