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.

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.

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

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

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

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.

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

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

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

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

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

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.

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

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

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

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

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

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

Latest Math Topics

Latest Math Problems

A best free mathematics education website for students, teachers and researchers.

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Learn how to solve the maths problems in different methods with understandable steps.

Copyright © 2012 - 2021 Math Doubts, All Rights Reserved