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

The irrational form of square root function can be removed by choosing the square of a variable as a replacement for variable $x$. So, let’s substitute square of a variable $u$ in the place of $x$.

$x \,=\, u^2$

Firstly, let’s find the integral of the $x$ with respect to $x$.

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

According to the derivative of a variable formula, the derivative of a variable $x$ with respect to $x$ is equal to one.

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

Now, let’s focus on finding the derivative of the $u$ square with respect to $x$.

The letter $u$ is a variable. So, its square function $u^2$ is not constant, whereas it is not possible to find the derivative of $u$ square with respect to $x$, but it is possible with chain rule.

The function $u^2$ should be differentiated with respect to $u$. So, let’s make an adjustment to find the derivative of $u^2$ with respect to $u$.

$\implies$ $1$ $\,=\,$ $\dfrac{d}{dx} \times 1 \times u^2$

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

The numerators of the fractions can be interchanged as per the multiplication of fractions.

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

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

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

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

Now, let’s find the derivative of $u$ square with respect to $u$ by the power rule of derivatives.

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

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

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

Now, let’s simplify the equation by removing the rational form.

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

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

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

$\implies$ $\dfrac{1}{1}$ $\,=\,$ $\dfrac{2udu}{dx}$

Finally, use the cross multiply rule to remove the rational form from the equation.

$\implies$ $1 \times dx$ $\,=\,$ $2udu \times 1$

$\,\,\,\,\,\,\therefore\,\,\,$ $dx \,=\, 2udu$

We have taken that $x \,=\, u^2$ and derived that $dx \,=\, 2udu$. Now, substitute them to convert a function in terms of $x$ into a function in terms of $u$.

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

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

Now, let’s focus on simplifying the function in the integration.

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

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

The first factor is a rational function. So, let’s express the second factor of the whole function in rational form.

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

Now, multiply both rational functions to get their product.

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

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

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

It is time to separate the constants from the function to prepare the function for integration.

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

Now, use the multiplication of fractions to separate the constants from the rational function.

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

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

The number $2$ is a constant and it can be taken out from the integration as per the constant multiple rule of integrals.

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

The expression in the numerator is $u^2$ and the same expression is also there in the first term of the denominator. So, add one to $u$ square in the numerator and also subtract one from their sum.

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

Now, let’s split the rational function as the difference between two fractions as per the subtraction of the fractions.

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

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

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

Let’s use the subtraction of integration formula to find the integral of difference between functions by evaluating the difference between their integrals.

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

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

Now, use the integral of one formula to find the integral of one with respect to $u$.

$\,\,=\,$ $2 \times \displaystyle \bigg(u+c_1-\int{\dfrac{1}{1+u^2}\bigg)}\,du$

The integral of one by one plus square of variable formula to find the integration of one divided by one plus $u$ square with respect to $u$.

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

In this integral problem, $c_1$ and $c_2$ are integral constants and let’s simplify the expression mathematically.

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

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

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

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

In this integration problem, $c_1$ and $c_2$ are constants, their difference is also a constant and the product is also a constant when the difference of constants is multiplied by the number $2$. So, let’s denote the integral constant by a letter $c$ simply.

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

We have taken that $x \,=\, u^2$. Therefore, $u \,=\, \sqrt{x}$. Now, substitute the value of $u$ by its equivalent value in the expression to get the integral of the given irrational function in rational form.

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