The indefinite integral of the quotient of one divided by the square root of $x$ plus two minus square root of $x$ plus one with respect to $x$ should be evaluated in this integration problem.

The quotient of 1 divided by the square root of x plus 2 minus the square root of x plus 1 is a reciprocal function basically. So, it is a rational function but due to the involvement of the square roots in the denominator, the rational function 1 divided by the square root of x plus 2 minus the square root of x plus 1 is an irrational function.

The function in the denominator is in irrational form and it creates problems while finding the integral of this irrational function. In this problem, the difference between the functions under the square roots is a number. Hence, the irrational form can be removed by rationalizing the function by its conjugate.

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

The conjugate of square root of $x$ plus $2$ minus the square root of $x$ plus $1$ is the square root of $x$ plus $2$ plus the square root of $x$ plus $1$. Therefore, factor one can be written in terms of conjugate function.

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

Now, two functions in fraction form are involved in multiplication and let’s find their product by using the multiplication of the fractions.

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

There are two factors in the denominator and compare them. Each factor is a binomial, and their terms are also same, but they have opposite signs. So, the difference of squares identity can be used to find the product of the binomials.

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

Look at the denominator, there are two functions in square root form and each function in radical form has square. So, the square eliminates the square root from the function.

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

The minus sign between the functions expresses that the difference between them should be evaluated to simplify the expression further in the denominator.

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

The places of the terms in the expression can be changed by using the commutative property.

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

There are two $x$ terms in the denominator and their difference is equal to zero. Similarly, the number one can be subtracted from the number two to find their difference.

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

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

$=\,\,$ $\displaystyle \int{\big(\sqrt{x+2}+\sqrt{x+1}\big)}\,dx$

The function is a sum of two irrational functions and the integral of sum of two functions can be evaluated by the sum of their integrals, as per the sum rule of the integration.

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

The $x$ plus two is a linear function in one variable in the first term and the $x$ plus one is also a linear function in one variable in the second term. The integrals of them can be evaluated but they are under control of square root. So, let’s first eliminate the square root by writing it in fraction form.

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

Now, the power rule of integration can be used to find the integral of every power function, but the binomial is there instead of a monomial at base position in every term. Hence, let’s denote the linear functions $x+2$ and $x+1$ by the variables.

Firstly, let’s denote the linear function $x+2$ by a variable $u$. Therefore, $u \,=\, x+2$. Now, differentiate both sides of the equation with respect to $x$.

$\implies$ $\dfrac{d}{dx}{\,u}$ $\,=\,$ $\dfrac{d}{dx}{(x+2)}$

According to the sum rule of the differentiation, the derivative of sum of terms $x$ and $2$ can be evaluated by the sum of their derivatives.

$\implies$ $\dfrac{d}{dx}{\,u}$ $\,=\,$ $\dfrac{d}{dx}{(x)}$ $+$ $\dfrac{d}{dx}{(2)}$

The derivative of $u$ cannot be evaluated with respect to $x$. So, write it as derivative of $u$ with respect to $x$. The derivative of $x$ with respect to $x$ is equal to one as per the derivative rule of a variable, and the derivative of two with respect to $x$ is equal to zero as per the derivative rule of a constant.

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

Now, let’s simplify the differential equation to find the differential element in $x$ in terms of the differential element in $u$.

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

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

$\implies$ $du$ $\,=\,$ $dx$

$\,\,\,\therefore\,\,\,\,\,\,$ $dx$ $\,=\,$ $du$

We have considered a literal $u$ to denote the linear function $x+2$ and it is derived by the differentiation that the differential element in $x$ is equal to the differential element in $u$. So, the integral of function in the first term can be expressed in terms of $u$ completely.

$\implies$ $\displaystyle \int{(x+2)^{\Large \frac{1}{2}}}\,dx$ $+$ $\displaystyle \int{(x+1)^{\Large \frac{1}{2}}}\,dx$ $\,=\,$ $\displaystyle \int{(u)^{\Large \frac{1}{2}}}\,du$ $+$ $\displaystyle \int{(x+1)^{\Large \frac{1}{2}}}\,dx$

$\,\,\,\therefore\,\,\,\,\,\,$ $\displaystyle \int{(x+2)^{\Large \frac{1}{2}}}\,dx$ $+$ $\displaystyle \int{(x+1)^{\Large \frac{1}{2}}}\,dx$ $\,=\,$ $\displaystyle \int{u^{\Large \frac{1}{2}}}\,du$ $+$ $\displaystyle \int{(x+1)^{\Large \frac{1}{2}}}\,dx$

Similarly, let’s denote the linear function $x+1$ by a variable $v$. So, $v \,=\, x+1$. Now, differentiate both sides of the equation with respect to $x$.

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

Now, use the addition rule of the derivatives to find the derivative of the linear function $x$ plus one by finding the sum of their derivatives.

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

The derivative of $v$ cannot be evaluated with respect to $x$. So, it written as the derivative of $v$ with respect to $x$ on the left hand side of the equation. According to the derivative rule of a variable, the derivative of $x$ with respect to $x$ is equal to one, and the derivative of one with respect to $x$ is equal to zero as per the derivative rule of a constant.

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

Let’s simplify the differential equation to find the differential element in $x$ in terms of the differential element in $v$.

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

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

$\implies$ $dv$ $\,=\,$ $dx$

$\,\,\,\therefore\,\,\,\,\,\,$ $dx$ $\,=\,$ $dv$

We have considered that the $x$ plus $1$ is denoted by a variable $v$ and it is evaluated that the differential element $dx$ is equal to the differential element $dv$ in the case of second term.

$\implies$ $\displaystyle \int{(x+2)^{\Large \frac{1}{2}}}\,dx$ $+$ $\displaystyle \int{(x+1)^{\Large \frac{1}{2}}}\,dx$ $\,=\,$ $\displaystyle \int{u^{\Large \frac{1}{2}}}\,du$ $+$ $\displaystyle \int{(v)^{\Large \frac{1}{2}}}\,dv$

$\,\,\,\therefore\,\,\,\,\,\,$ $\displaystyle \int{(x+2)^{\Large \frac{1}{2}}}\,dx$ $+$ $\displaystyle \int{(x+1)^{\Large \frac{1}{2}}}\,dx$ $\,=\,$ $\displaystyle \int{u^{\Large \frac{1}{2}}}\,du$ $+$ $\displaystyle \int{v^{\Large \frac{1}{2}}}\,dv$

The integral equation expresses that the left hand side of the equation can be evaluated by finding the right hand side of the equation.

$\displaystyle \int{u^{\Large \frac{1}{2}}}\,du$ $+$ $\displaystyle \int{v^{\Large \frac{1}{2}}}\,dv$

The power rule of integration can be used to find the integral of every function in the expression. The $c_1$ and $c_2$ are used here to express the constants of integration.

$=\,\,\,$ $\dfrac{u^{\Large \frac{1}{2} \normalsize +1}}{\dfrac{1}{2}+1}$ $+$ $c_1$ $+$ $\dfrac{v^{\Large \frac{1}{2} \normalsize +1}}{\dfrac{1}{2}+1}$ $+$ $c_2$

The commutative property of addition can be used to change the places of the terms in the expression.

$=\,\,\,$ $\dfrac{u^{\Large \frac{1}{2} \normalsize +1}}{\dfrac{1}{2}+1}$ $+$ $\dfrac{v^{\Large \frac{1}{2} \normalsize +1}}{\dfrac{1}{2}+1}$ $+$ $c_1$ $+$ $c_2$

In this expression, the $c_1$ and $c_2$ are the constants and their sum is also a constant. Hence, the sum of them can be denoted by a constant $c$ simply.

$=\,\,\,$ $\dfrac{u^{\Large \frac{1}{2} \normalsize +1}}{\dfrac{1}{2}+1}$ $+$ $\dfrac{v^{\Large \frac{1}{2} \normalsize +1}}{\dfrac{1}{2}+1}$ $+$ $c$

Look at the expression at exponent position in numerator and the expression in the denominator of the first and second terms. The numerator one is added to the quotient of one divided by two and their sum can be evaluated by the addition of the fractions.

$=\,\,\,$ $\dfrac{u^{\Large \frac{1+2}{2}}}{\dfrac{1+2}{2}}$ $+$ $\dfrac{v^{\Large \frac{1+2}{2}}}{\dfrac{1+2}{2}}$ $+$ $c$

$=\,\,\,$ $\dfrac{u^{\Large \frac{3}{2}}}{\dfrac{3}{2}}$ $+$ $\dfrac{v^{\Large \frac{3}{2}}}{\dfrac{3}{2}}$ $+$ $c$

There is a rational number in the denominator of both first and second terms and they can be separated from the fractions by using the multiplication of the fractions.

$=\,\,\,$ $\dfrac{1 \times u^{\Large \frac{3}{2}}}{\dfrac{3}{2}}$ $+$ $\dfrac{1 \times v^{\Large \frac{3}{2}}}{\dfrac{3}{2}}$ $+$ $c$

$=\,\,\,$ $\dfrac{1}{\dfrac{3}{2}} \times u^{\Large \frac{3}{2}}$ $+$ $\dfrac{1}{\dfrac{3}{2}} \times v^{\Large \frac{3}{2}}$ $+$ $c$

Now, use the reciprocal of fractions rule to find the reciprocal of a rational number.

$=\,\,\,$ $\dfrac{2}{3} \times u^{\Large \frac{3}{2}}$ $+$ $\dfrac{2}{3} \times v^{\Large \frac{3}{2}}$ $+$ $c$

$=\,\,\,$ $\dfrac{2}{3}u^{\Large \frac{3}{2}}$ $+$ $\dfrac{2}{3}v^{\Large \frac{3}{2}}$ $+$ $c$

$=\,\,\,$ $\dfrac{2}{3}u^{\Large \frac{3}{2}}$ $+$ $\dfrac{2}{3}v^{\Large \frac{3}{2}}$ $+$ $c$

$=\,\,\,$ $\dfrac{2}{3}u^{\Large \frac{3}{2}}$ $+$ $\dfrac{2}{3}v^{\Large \frac{3}{2}}$ $+$ $c$

$=\,\,\,$ $\dfrac{2}{3}(x+2)^{\Large \frac{3}{2}}$ $+$ $\dfrac{2}{3}(x+1)^{\Large \frac{3}{2}}$ $+$ $c$

Latest Math Topics

Dec 13, 2023

Jul 20, 2023

Jun 26, 2023

Latest Math Problems

Jan 30, 2024

Oct 15, 2023

Copyright © 2012 - 2023 Math Doubts, All Rights Reserved