The one over the square root of $x$ plus one minus fourth root of $x$ plus one is a given rational function but two functions in root form are involved in forming this rational form. Due to the involvement of square root and fourth root in the denominator, the rational expression is known as an irrational function.

The indefinite integral of the one divided by the square root of $x$ plus one minus the fourth root of $x$ plus one should be evaluated in this integration problem. So, let’s learn how to find the integral of given irrational function in $x$.

The roots in the irrational functions in the denominator can be expressed in the exponential notation by writing them as the rational numbers.

$\implies$ $\displaystyle \int{\dfrac{1}{\sqrt{x+1}-\sqrt[\Large 4]{x+1}}}\,dx$ $\,=\,$ $\displaystyle \int{\dfrac{1}{(x+1)^{\Large \frac{1}{2}}-(x+1)^{\Large \frac{1}{4}}}}\,dx$

The base of both functions in the denominator are same and it can be denoted by a variable $u$ but it does not remove the irrational form from the functions in the denominator.

$\implies$ $(x+1)^{\Large \frac{1}{2}}-(x+1)^{\Large \frac{1}{4}}$ $\,=\,$ $u^{\Large \frac{1}{2}}-u^{\Large \frac{1}{4}}$

So, a suitable substitution should be taken to denote the function in base position and also to eliminate the rational powers from the exponent position in the denominator.

Now, let’s try to denote the expression $x$ plus one by the square of a variable $v$.

$\implies$ $(x+1)^{\Large \frac{1}{2}}-(x+1)^{\Large \frac{1}{4}}$ $\,=\,$ $(v^2)^{\Large \frac{1}{2}}-(v^2)^{\Large \frac{1}{4}}$ $\,=\,$ $v-v^{\Large \frac{1}{2}}$

It eliminated the irrational form from the first function but it is failed to eliminate the irrational form from the second function. So, it is proved that the square of variable $v$ is not a suitable substitution of the expression $x$ plus one.

Now, let’s denote the expression $x$ plus one by the fourth power of a variable $y$.

$\implies$ $(x+1)^{\Large \frac{1}{2}}-(x+1)^{\Large \frac{1}{4}}$ $\,=\,$ $(y^4)^{\Large \frac{1}{2}}-(y^4)^{\Large \frac{1}{4}}$ $\,=\,$ $y^2-y$

The variable $y$ to the fourth power eliminated the irrational form successfully from the both functions in the denominator. So, the variable $y$ raised to the fourth power is a suitable substitution to rationalize the irrational function.

$y^4$ $\,=\,$ $x+1$

Now, differentiate the expressions on both sides of the equation with respect to $x$.

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

On the left hand side of the equation, $y$ to the fourth power is a power function and its derivative can be evaluated by the power rule of the differentiation.

$\implies$ $4y^{4-1}\dfrac{dy}{dx}$ $\,=\,$ $\dfrac{d}{dx}{(x+1)}$

$\implies$ $4y^3\dfrac{dy}{dx}$ $\,=\,$ $\dfrac{d}{dx}{(x+1)}$

On the right hand side of the equation, the derivative of sum of two functions can be evaluated by using the sum rule of the differentiation.

$\implies$ $4y^3\dfrac{dy}{dx}$ $\,=\,$ $\dfrac{d}{dx}{(x)}$ $+$ $\dfrac{d}{dx}{(1)}$

Now, use the derivative rule of a variable to find the derivative of a variable $x$ with respect to $x$ and also use the derivative rule of a constant to find the derivative of one with respect to $x$.

$\implies$ $4y^3\dfrac{dy}{dx}$ $\,=\,$ $1$ $+$ $0$

Now, let’s find the differential element $dx$ by simplifying the differential equation.

$\implies$ $4y^3\dfrac{dy}{dx}$ $\,=\,$ $1$

$\implies$ $4y^3\,dy$ $\,=\,$ $1 \times dx$

$\implies$ $4y^3\,dy$ $\,=\,$ $dx$

$\,\,\,\therefore\,\,\,\,\,\,$ $dx$ $\,=\,$ $4y^3\,dy$

The expression $x$ plus one is denoted by $y$ raised to the power of four and the differential element $dx$ is also calculated in terms of $y$. So, it is time to substitute them in the function in terms of $x$ and it transforms the whole function in terms of $y$.

$\implies$ $\displaystyle \int{\dfrac{1}{(x+1)^{\Large \frac{1}{2}}-(x+1)^{\Large \frac{1}{4}}}}\,dx$ $\,=\,$ $\displaystyle \int{\dfrac{1}{(y^4)^{\Large \frac{1}{2}}-(y^4)^{\Large \frac{1}{4}}}}\,(4y^3dy)$

Let’s focus on simplifying the function.

$=\,\,$ $\displaystyle \int{\dfrac{1}{(y^4)^{\Large \frac{1}{2}}-(y^4)^{\Large \frac{1}{4}}}}\,(4y^3dy)$

Firstly, let’s split the second factor into two factors to separate the differential element $dy$ from the function.

$=\,\,$ $\displaystyle \int{\dfrac{1}{(y^4)^{\Large \frac{1}{2}}-(y^4)^{\Large \frac{1}{4}}}}\,\times 4y^3 \times dy$

The first factor is a rational function and the second factor is not a fraction. However, they can be multiplied by the multiplication of the fractions.

$=\,\,$ $\displaystyle \int{\dfrac{1\times 4y^3}{(y^4)^{\Large \frac{1}{2}}-(y^4)^{\Large \frac{1}{4}}}}\,\times dy$

$=\,\,$ $\displaystyle \int{\dfrac{4y^3}{(y^4)^{\Large \frac{1}{2}}-(y^4)^{\Large \frac{1}{4}}}}\,dy$

$=\,\,$ $\displaystyle \int{\dfrac{4y^3}{(y^4)^{\Large \frac{1}{2}}-(y^4)^{\Large \frac{1}{4}}}}\,dy$

Now, let’s concentrate on the expression in the denominator. There is fraction at exponent position of a power function in both terms and each term can be simplified by the power rule of exponents.

$=\,\,$ $\displaystyle \int{\dfrac{4y^3}{y^{4 \times \Large \frac{1}{2}}-y^{4 \times \Large \frac{1}{4}}}}\,dy$

$=\,\,$ $\displaystyle \int{\dfrac{4y^3}{y^{\Large \frac{4 \times 1}{2}}-y^{\Large \frac{4 \times 1}{4}}}}\,dy$

$=\,\,$ $\displaystyle \int{\dfrac{4y^3}{y^{\Large \frac{4}{2}}-y^{\Large \frac{4}{4}}}}\,dy$

$=\,\,$ $\displaystyle \int{\dfrac{4y^3}{y^{\Large \frac{\cancel{4}}{\cancel{2}}}-y^{\Large \frac{\cancel{4}}{\cancel{4}}}}}\,dy$

$=\,\,$ $\displaystyle \int{\dfrac{4y^3}{y^2-y^1}}\,dy$

$=\,\,$ $\displaystyle \int{\dfrac{4y^3}{y^2-y}}\,dy$

There is a variable $y$ commonly in both terms in the denominator. So, the common factor can be taken out from the both terms in the denominator.

$=\,\,$ $\displaystyle \int{\dfrac{4y^3}{y \times y-y \times 1}}\,dy$

$=\,\,$ $\displaystyle \int{\dfrac{4y^3}{y \times (y-1)}}\,dy$

$=\,\,$ $\displaystyle \int{\dfrac{4\cancel{y^3}}{\cancel{y} \times (y-1)}}\,dy$

$=\,\,$ $\displaystyle \int{\dfrac{4y^2}{y-1}}\,dy$

The number $4$ is a constant and it can be released from the integral operation. So, let’s split the rational function as a product of two functions as per the multiplication of the fractions.

$=\,\,$ $\displaystyle \int{\dfrac{4 \times y^2}{y-1}}\,dy$

$=\,\,$ $\displaystyle \int{\bigg(4 \times \dfrac{y^2}{y-1}\bigg)}\,dy$

The constant factor $4$ can be released from the integration by the constant multiple rule of integration.

$=\,\,$ $4 \times \displaystyle \int{\dfrac{y^2}{y-1}}\,dy$

The expressions in both numerator and denominator are polynomials. The degree of the polynomial in the numerator is two and the degree of polynomial in the denominator is one. It means, the degree of the polynomial in the numerator is greater than the degree of the polynomial in the denominator. So, the polynomial in the numerator should be divided by the polynomial in the denominator.

Now, subtract one from the expression in the numerator and add one to their difference to include a number one properly.

$=\,\,$ $4 \times \displaystyle \int{\dfrac{y^2-1+1}{y-1}}\,dy$

Now, split the rational function as the sum of two fractions as per the addition rule of the fractions.

$=\,\,$ $4 \times \displaystyle \int{\bigg(\dfrac{y^2-1}{y-1}-\dfrac{1}{y-1}\bigg)}\,dy$

Look at the numerator of the first term in the expression inside the integration. The first term in the numerator is in square form and the number one can be written as square of one to convert the expression as the difference of two quantities.

$=\,\,$ $4 \times \displaystyle \int{\bigg(\dfrac{y^2-1^2}{y-1}-\dfrac{1}{y-1}\bigg)}\,dy$

According to the difference of squares formula, the difference of quantities in square form can be written in factor form and it helps us to eliminate a factor by the expression in the denominator.

$=\,\,$ $4 \times \displaystyle \int{\bigg(\dfrac{(y+1)(y-1)}{y-1}-\dfrac{1}{y-1}\bigg)}\,dy$

$=\,\,$ $4 \times \displaystyle \int{\bigg(\dfrac{(y+1)\cancel{(y-1)}}{\cancel{y-1}}-\dfrac{1}{y-1}\bigg)}\,dy$

$=\,\,$ $4 \times \displaystyle \int{\bigg((y+1)-\dfrac{1}{y-1}\bigg)}\,dy$

$=\,\,$ $4 \times \displaystyle \int{\bigg(y+1-\dfrac{1}{y-1}\bigg)}\,dy$

Let’s analyze the expression inside the integral operation. The first two terms express the addition of the two functions and the third term with negative sign expresses the difference between two functions. The integral of this expression can be evaluated by using the combination of addition rule and subtraction rule of the integration.

$=\,\,$ $4 \times \bigg(\displaystyle \int{y}\,dy$ $+$ $\displaystyle \int{1}\,dy$ $-$ $\displaystyle \int{\dfrac{1}{y-1}}\,dy\bigg)$

Use the power rule of integration to find the integration of a variable $y$ with respect to $y$ and use integral rule of one to find the integration of one with respect to $y$. Finally, use the integration rule of linear expression in reciprocal form to find the integral of the reciprocal of linear expression $y$ minus one with respect to $y$.

$=\,\,$ $4 \times \bigg(\dfrac{y^2}{2}+c_1$ $+$ $y+c_2$ $-$ $\log_{e}{|y-1|}+c_3\bigg)$

Here, $c_1$, $c_2$ and $c_3$ are the constants of integration and let’s write the terms in the expression in an order.

$=\,\,$ $4 \times \bigg(\dfrac{y^2}{2}$ $+$ $y$ $-$ $\log_{e}{|y-1|}$ $+$ $c_1+c_2+c_3\bigg)$

$=\,\,$ $4 \times \bigg(\dfrac{y^2}{2}$ $+$ $y$ $-$ $\log_{e}{|y-1|}$ $+$ $(c_1+c_2+c_3)\bigg)$

The number four is a coefficient of all the terms in the expression and it belongs to all the terms. So, let’s multiply all the terms by the number four.

$=\,\,$ $4 \times \dfrac{y^2}{2}$ $+$ $4 \times y$ $-$ $4 \times \log_{e}{|y-1|}$ $+$ $4 \times (c_1+c_2+c_3)$

$=\,\,$ $\dfrac{4 \times y^2}{2}$ $+$ $4y$ $-$ $4\log_{e}{|y-1|}$ $+$ $4(c_1+c_2+c_3)$

$=\,\,$ $\dfrac{\cancel{4} \times y^2}{\cancel{2}}$ $+$ $4y$ $-$ $4\log_{e}{|y-1|}$ $+$ $4(c_1+c_2+c_3)$

$=\,\,$ $2 \times y^2$ $+$ $4y$ $-$ $4\log_{e}{|y-1|}$ $+$ $4(c_1+c_2+c_3)$

$=\,\,$ $2y^2$ $+$ $4y$ $-$ $4\log_{e}{|y-1|}$ $+$ $4(c_1+c_2+c_3)$

The four times sum of the constants $c_1,$ $c_2$ and $c_3$ denotes a constant. So, it can represented by the constant of integration $c$. Now, the number four is a common factor in both second and third terms and It can be taken out common from them to express the whole expression in simple form.

$=\,\,$ $2y^2$ $+$ $4(y-\log_{e}{|y-1|})$ $+$ $c$

We have considered that $y^4$ $\,=\,$ $x+1$

$\implies$ $y$ $\,=\,$ $(x+1)^{\Large \frac{1}{4}}$

Now, it is time to replace the variable $y$ by its actual value to get the solution in terms of $x$.

$=\,\,$ $2\Big((x+1)^{\Large \frac{1}{4}}\Big)^2$ $+$ $4\Big((x+1)^{\Large \frac{1}{4}}-\log_{e}{\Big|(x+1)^{\Large \frac{1}{4}}-1\Big|}\Big)$ $+$ $c$

Use the power rule of exponents to simplify the first term and the rational number at exponent position in the second term can be written in root form.

$=\,\,$ $2(x+1)^{{\Large \frac{1}{4}} \times 2}$ $+$ $4\Big(\sqrt[\Large 4]{x+1}-\log_{e}{\big|\sqrt[\Large 4]{x+1}-1\big|}\Big)$ $+$ $c$

$=\,\,$ $2(x+1)^{\Large \frac{1 \times 2}{4}}$ $+$ $4\Big(\sqrt[\Large 4]{x+1}-\log_{e}{\big|\sqrt[\Large 4]{x+1}-1\big|}\Big)$ $+$ $c$

$=\,\,$ $2(x+1)^{\Large \frac{2}{4}}$ $+$ $4\Big(\sqrt[\Large 4]{x+1}-\log_{e}{\big|\sqrt[\Large 4]{x+1}-1\big|}\Big)$ $+$ $c$

$=\,\,$ $2(x+1)^{\Large \frac{\cancel{2}}{\cancel{4}}}$ $+$ $4\Big(\sqrt[\Large 4]{x+1}-\log_{e}{\big|\sqrt[\Large 4]{x+1}-1\big|}\Big)$ $+$ $c$

$=\,\,$ $2(x+1)^{\Large \frac{1}{2}}$ $+$ $4\Big(\sqrt[\Large 4]{x+1}-\log_{e}{\big|\sqrt[\Large 4]{x+1}-1\big|}\Big)$ $+$ $c$

Finally, write the rational number at exponent position in the first term in root form to finish the process of simplifying the expression.

$=\,\,$ $2\sqrt{x+1}$ $+$ $4\Big(\sqrt[\Large 4]{x+1}-\log_{e}{\big|\sqrt[\Large 4]{x+1}-1\big|}\Big)$ $+$ $c$

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