# Evaluate $\displaystyle \int{\dfrac{\sqrt{x}}{x+1}}\,dx$

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

The $\sqrt{x}$ divided by $x+1$ is basically a rational function but due to the involvement of a root function in the numerator, the rational function is called an irrational function. Now, let’s learn how to find the indefinite integration of an irrational function the square root of $x$ divided by $x$ plus $1$ with respect to $x$.

### Try Rationalizing substitution to remove radical form

The expression in the numerator $\sqrt{x}$ is an irrational function and the expression in the denominator $x+1$ is a linear expression in one variable in this function. The irrational function in the numerator creates hurdles while finding the integration of the given irrational function in rational form.

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

So, it is better to eliminate the radical form from the function by a suitable substitute and it helps us to convert the irrational function in rational form as a rational function purely. Now, let’s use the integration by rationalizing substitution method.

Let’s denote the square root of $x$ by a variable $u$.

$u \,=\, \sqrt{x}$

Take square on both sides of the equation to remove the root from the equation.

$\implies$ $u^2 = \big(\sqrt{x}\big)^2$

$\,\,\,\therefore\,\,\,\,\,\,$ $u^2 = x$

Now, differentiate the equation $u$ square equals to $x$ with respect to $x$ to convert the given irrational function as a rational function purely.

$\implies$ $\dfrac{d}{dx}\,u^2$ $\,=\,$ $\dfrac{d}{dx}\,x$

On the left hand side of the equation, the function is expressed in terms of $u$ but it should be differentiated with respect to $x$. However, it can be differentiated with respect to $x$ by using the chain rule. On the right hand side of the equation, the derivative of $x$ with respect to $x$ is equal to $1$ as per the derivative rule of a variable.

$\implies$ $2u\dfrac{du}{dx}$ $\,=\,$ $1$

Now, let’s find the value of the differential element in $x$ by cross multiplication.

$\implies$ $2u\,du$ $\,=\,$ $1 \times dx$

$\implies$ $2u\,du$ $\,=\,$ $dx$

$\,\,\,\,\,\,\therefore\,\,\,$ $dx$ $\,=\,$ $2u\,du$

The square root of $x$ is denoted by a variable $u$ and it is evaluated from this equation that the differential element $dx$ is equal to two times the product of variable $u$ and different element $du$. Now, the integral of a function in $x$ can be converted into the integral of a function in terms of $u$ by the two equations.

$\implies$ $\displaystyle \int{\dfrac{\sqrt{x}}{x+1}}\,dx$ $\,=\,$ $\displaystyle \int{\dfrac{u}{u^2+1}}\,2u\,du$

The equation expresses that the indefinite integral of the square root of $x$ divided by $x$ plus $1$ with respect to $x$ can be evaluated by finding the indefinite integral of the product of $u$ divided by $u$ square plus $1$ and, $2$ times $u$ with respect to $u$.

### Preparing the function for Integration by simplification

The quotient of $u$ divided by $u$ square plus $1$ and $2u$ are two factors and the integral of their product should be evaluated with respect to $u$. The $u$ divided by $u$ square plus $1$ is a fraction and $2u$ is not a fraction. However, they can be multiplied by the multiplication of the fractions.

$\,=\,$ $\displaystyle \int{\dfrac{u \times 2u}{u^2+1}}\,du$

$\,=\,$ $\displaystyle \int{\dfrac{2u^2}{u^2+1}}\,du$

The coefficient of $u$ square in the numerator is a constant. So, it can be released from the integral operation.

$\,=\,$ $\displaystyle \int{\dfrac{2 \times u^2}{u^2+1}}\,du$

$\,=\,$ $\displaystyle \int{\bigg(2 \times \dfrac{u^2}{u^2+1}\bigg)}\,du$

Now, use the constant multiple rule of integration to separate the factor $2$ from the integration.

$\,=\,$ $2 \times \displaystyle \int{\dfrac{u^2}{u^2+1}}\,du$

Look at the expressions in the both numerator and denominator of the rational function. The expressions in both numerator and denominator are second degree polynomial. So, the expression in the numerator can be divided by the expression in the denominator.

The second term in the denominator is $1$. So, add one to the expression in the numerator and subtract the $1$ from their sum to balance the inclusion.

$\,=\,$ $2 \times \displaystyle \int{\dfrac{u^2+1-1}{u^2+1}}\,du$

Now, compare the expression in the denominator with the expression in the numerator. The expression $u$ square plus $1$ is also there in the numerator. So, the fraction can be split as the difference of two fractions as per the difference rule of fractions with same denominator.

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

Now, let’s try to simplify the expression by simplifying each term in the expression.

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

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

### Find the Indefinite integration of the function

The function inside the integral operation is in difference form. So, the indefinite integration of the difference of the functions should be evaluated with respect to $x$. It can be evaluated by the difference rule of the integration.

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

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

The indefinite integral of one with respect to $x$ can be evaluated by the integral rule of one. Similarly, the indefinite integral of $1$ divided by $1$ plus $x$ square can also be evaluated with respect to $x$ by the integral rule of sum of squares in reciprocal form.

$\,=\,$ $2 \times \big(u+c_1-(\arctan{u}+c_2)\big)$

or

$\,=\,$ $2 \times \big(u+c_1-(\tan^{-1}{u}+c_2)\big)$

The second factor is an expression in which $c_1$ and $c_2$ are the constants of integration. Now, let us simplify the expression by simplifying the expression in the second factor.

$\,=\,$ $2 \times (u+c_1-\arctan{u}-c_2)$

$\,=\,$ $2 \times (u-\arctan{u}+c_1-c_2)$

Now, consider the functions as a group and the constants as another group in the expression for our convenience.

$\,=\,$ $2 \times \big((u-\arctan{u})+(c_1-c_2)\big)$

The number $2$ multiplies both the terms in the expression. So, it can be distributed over the addition by the distributive property of the multiplication over the addition.

$\,=\,$ $2 \times (u-\arctan{u})$ $+$ $2 \times (c_1-c_2)$

$\,=\,$ $2(u-\arctan{u})$ $+$ $2(c_1-c_2)$

In the second term, $c_1$ and $c_2$ are the constants and their difference is also a constant. Similarly, the two times that constant is also a constant. So, it can be simply denoted by a constant $c$.

$\,=\,$ $2(u-\arctan{u})+c$

The variable $u$ is taken to denote the square root of $x$ in this integration problem. So, replace the variable $u$ by its actual value in the above equation to get the required solution.

$\,=\,$ $2\big(\sqrt{x}-\arctan{\sqrt{x}}\big)+c$

It is alternatively written as follows in mathematics.

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

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Jun 26, 2023

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