In this indefinite integral problem, square root of $x$ is multiplied by the $2$ plus $3$ times $x$ whole power of three by two. The two functions are irrational functions and the indefinite integration of the reciprocal of their product should be evaluated with respect to $x$ in this problem.

$\displaystyle \int{\frac{1}{x^{\Large \frac{1}{2}}(2+3x)^{\Large \frac{3}{2}}}\,}dx$

In this problem, the $x$ raised to the power of half does not create any issue for the integration, but the issue comes with the binomial $2+3x$ whole power of $3$ by $2$. The reciprocal property and the irrationality of the binomial create the problem for the indefinite integration. So, it is tough to evaluate the indefinite integration of these types of irrational functions.

In order to solve such issues, the binomial is recommendable to express as a monomial. The binomial contains a variable $x$. Due to this reason; the monomial should have the same variable. Hence, the value of $2+3x$ can be expressed as a product of two variables $x$ and $u$.

Take $ux \,=\, 2+3x$

Now, evaluate the value of $x$ in terms of $u$.

$\implies$ $ux-3x \,=\, 2$

$\implies$ $x(u-3) \,=\, 2$

$\,\,\,\therefore\,\,\,\,\,\,$ $x \,=\, \dfrac{2}{u-3}$

Now, find the value of irrational function by eliminating the $x$ with its equivalent value.

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

$\,\,\,=\,\,$ $x^{\Large \frac{1}{2}}(u \times x)^{\Large \frac{3}{2}}$

According to the power of a product rule, the second factor can be written as a product of two functions.

$\,\,\,=\,\,$ $x^{\Large \frac{1}{2}} \times u^{\Large \frac{3}{2}} \times x^{\Large \frac{3}{2}}$

$\,\,\,=\,\,$ $x^{\Large \frac{1}{2}} \times x^{\Large \frac{3}{2}} \times u^{\Large \frac{3}{2}}$

Now, use the product rule of exponents with same base to calculate the product of the first two factors in the expression.

$\,\,\,=\,\,$ $x^{\Large \frac{1}{2}\normalsize + \Large \frac{3}{2}} \times u^{\Large \frac{3}{2}}$

Add the fractions at exponent position of the first factor.

$\,\,\,=\,\,$ $x^{\Large \frac{1+3}{2}} \times u^{\Large \frac{3}{2}}$

$\,\,\,=\,\,$ $x^{\Large \frac{4}{2}} \times u^{\Large \frac{3}{2}}$

$\,\,\,=\,\,$ $x^{\Large \frac{\cancel{4}}{\cancel{2}}} \times u^{\Large \frac{3}{2}}$

$\,\,\,=\,\,$ $x^2 \times u^{\Large \frac{3}{2}}$

It is time to eliminate the $x$ by its value in terms of $u$.

$\,\,\,=\,\,$ $\bigg(\dfrac{2}{u-3}\bigg)^2 \times u^{\Large \frac{3}{2}}$

Now, use the power of a quotient rule to expand the first factor further.

$\,\,\,=\,\,$ $\dfrac{2^2}{(u-3)^2} \times u^{\Large \frac{3}{2}}$

$\,\,\,=\,\,$ $\dfrac{4}{(u-3)^2} \times u^{\Large \frac{3}{2}}$

Use the multiplication of fractions method to get the product of the factors.

$\,\,\,=\,\,$ $\dfrac{4\times u^{\Large \frac{3}{2}}}{(u-3)^2}$

$\implies$ $x^{\Large \frac{1}{2}}(2+3x)^{\Large \frac{3}{2}}$ $\,=\,$ $\dfrac{4u^{\Large \frac{3}{2}}}{(u-3)^2}$

Now, it is time to evaluate its reciprocal.

$\implies$ $\dfrac{1}{x^{\Large \frac{1}{2}}(2+3x)^{\Large \frac{3}{2}}}$ $\,=\,$ $\dfrac{1}{\dfrac{4u^{\Large \frac{3}{2}}}{(u-3)^2}}$

$\,\,\,\therefore\,\,\,\,\,\,$ $\dfrac{1}{x^{\Large \frac{1}{2}}(2+3x)^{\Large \frac{3}{2}}}$ $\,=\,$ $\dfrac{(u-3)^2}{4u^{\Large \frac{3}{2}}}$

The value of given irrational function in reciprocal form is expressed in terms of $u$.

The irrational function is successfully converted in terms of $u$. Similarly, the differential $dx$ should also be converted in terms of $u$.

We have taken that $ux \,=\, 2+3x$

Now, differentiate the mathematical equation with respect to $x$.

$\implies$ $\dfrac{d}{dx}{(ux)}$ $\,=\,$ $\dfrac{d}{dx}{(2+3x)}$

The derivative of product of two functions can be calculated by the product rule of differentiation. Similarly, the derivative of sum of two functions can also be calculated by the sum rule of differentiation.

$\implies$ $u \times \dfrac{dx}{dx}$ $+$ $x \times \dfrac{du}{dx}$ $\,=\,$ $\dfrac{d}{dx}{(2)}$ $+$ $\dfrac{d}{dx}{(3x)}$

According to the derivative rule of variable, the derivative of a variable with respect to same variable is one. The derivative rule of a constant, the derivative of a constant is zero.

$\implies$ $u \times 1$ $+$ $x \times \dfrac{du}{dx}$ $\,=\,$ $0$ $+$ $\dfrac{d}{dx}{(3 \times x)}$

Now, use the constant multiple rule of the differentiation to release the constant factor $3$ from the differentiation.

$\implies$ $u$ $+$ $x \times \dfrac{du}{dx}$ $\,=\,$ $3 \times \dfrac{d}{dx}{(x)}$

$\implies$ $u$ $+$ $x \times \dfrac{du}{dx}$ $\,=\,$ $3 \times \dfrac{dx}{dx}$

$\implies$ $u$ $+$ $x \times \dfrac{du}{dx}$ $\,=\,$ $3 \times 1$

$\implies$ $u$ $+$ $x \times \dfrac{du}{dx}$ $\,=\,$ $3$

Now, let’s simplify this equation to find the value of differential $dx$.

$\implies$ $x \times \dfrac{du}{dx}$ $\,=\,$ $3-u$

$\implies$ $\dfrac{x}{3-u} \times du$ $\,=\,$ $dx$

$\implies$ $dx$ $\,=\,$ $\dfrac{x}{3-u} \times du$

$\implies$ $dx$ $\,=\,$ $\dfrac{\dfrac{2}{u-3}}{3-u} \times du$

$\implies$ $dx$ $\,=\,$ $\dfrac{2}{(u-3)(3-u)} \times du$

$\implies$ $dx$ $\,=\,$ $\dfrac{2}{(3-u)(u-3)} \times du$

$\implies$ $dx$ $\,=\,$ $\dfrac{2}{-(u-3)(u-3)} \times du$

$\implies$ $dx$ $\,=\,$ $\dfrac{2}{-(u-3)^2} \times du$

$\,\,\,\therefore\,\,\,\,\,\,$ $dx$ $\,=\,$ $\Bigg(\dfrac{-2}{(u-3)^2}\Bigg) \times du$

In the first step, we have evaluated the value of the given irrational function in terms of $u$. Similarly, in the above step, the value of the differential $dx$ is also evaluated in terms of $u$.

$(1).\,\,\,$ $\dfrac{1}{x^{\Large \frac{1}{2}}(2+3x)^{\Large \frac{3}{2}}}$ $\,=\,$ $\dfrac{(u-3)^2}{4u^{\Large \frac{3}{2}}}$

$(2).\,\,\,$ $dx$ $\,=\,$ $\Bigg(\dfrac{-2}{(u-3)^2}\Bigg) \times du$

Now, substitute them in the given indefinite integration problem to start finding the integral of the given irrational function with respect to $x$.

$\implies$ $\displaystyle \int{\frac{1}{x^{\Large \frac{1}{2}}(2+3x)^{\Large \frac{3}{2}}}\,}dx$ $\,=\,$ $\displaystyle \int{\dfrac{(u-3)^2}{4u^{\Large \frac{3}{2}}}} \times \Bigg(\dfrac{-2}{(u-3)^2}\Bigg) \times du$

This equation clears that the integration of the given irrational function can be evaluated by calculating the integration of the function in terms of $u$. Firstly, let’s simplify the product of the functions in terms of $u$ before implementing the procedure of the integration. So, get the product of the functions in fraction form by multiplication.

$\,\,\,=\,$ $\displaystyle \int{\dfrac{(u-3)^2 \times (-2)}{4u^{\Large \frac{3}{2}} \times (u-3)^2}\,} du$

$\,\,\,=\,\,$ $\displaystyle \int{\dfrac{\cancel{(u-3)^2} \times (-2)}{4u^{\Large \frac{3}{2}} \times \cancel{(u-3)^2}}\,} du$

$\,\,\,=\,\,$ $\displaystyle \int{\dfrac{(-2)}{4u^{\Large \frac{3}{2}}}\,} du$

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

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

$\,\,\,=\,\,$ $\displaystyle \int{\dfrac{(-1) \times 1}{2 \times u^{\Large \frac{3}{2}}}\,} du$

$\,\,\,=\,\,$ $\displaystyle \int{\Bigg(\dfrac{(-1)}{2} \times \dfrac{1}{u^{\Large \frac{3}{2}}}\Bigg)\,} du$

The constant factor can be separated from the integration as per the constant multiple rule of the integration.

$\,\,\,=\,\,$ $\dfrac{(-1)}{2} \times \displaystyle \int{\dfrac{1}{u^{\Large \frac{3}{2}}}\,} du$

$\,\,\,=\,\,$ $-\dfrac{1}{2} \times \displaystyle \int{\dfrac{1}{u^{\Large \frac{3}{2}}}\,} du$

$\,\,\,=\,\,$ $-\dfrac{1}{2} \times \displaystyle \int{u^{-\Large \frac{3}{2}}\,} du$

The integration of the function can be evaluated as per the power rule of integration.

$\,\,\,=\,\,$ $-\dfrac{1}{2} \times \Bigg(\dfrac{u^{-\Large \frac{3}{2}\normalsize +1}}{-\dfrac{3}{2}+1}+c_1\Bigg)$

$\,\,\,=\,\,$ $-\dfrac{1}{2} \times \Bigg(\dfrac{u^{\Large \frac{-3+2}{2}}}{\dfrac{-3+2}{2}}+c_1\Bigg)$

$\,\,\,=\,\,$ $-\dfrac{1}{2} \times \Bigg(\dfrac{u^{\Large \frac{-1}{2}}}{\dfrac{-1}{2}}+c_1\Bigg)$

$\,\,\,=\,\,$ $-\dfrac{1}{2} \times \Bigg(\dfrac{u^{-\Large \frac{1}{2}}}{-\dfrac{1}{2}}+c_1\Bigg)$

$\,\,\,=\,\,$ $-\dfrac{1}{2} \times \Bigg(\dfrac{1 \times u^{-\Large \frac{1}{2}}}{-\dfrac{1}{2}}+c_1\Bigg)$

$\,\,\,=\,\,$ $-\dfrac{1}{2} \times \Bigg(\dfrac{1}{-\dfrac{1}{2}} \times u^{-\Large \frac{1}{2}}+c_1\Bigg)$

$\,\,\,=\,\,$ $-\dfrac{1}{2} \times \Bigg((-2) \times u^{-\Large \frac{1}{2}}+c_1\Bigg)$

$\,\,\,=\,\,$ $-\dfrac{1}{2} \times (-2) \times u^{-\Large \frac{1}{2}}-\dfrac{1}{2} \times c_1$

$\,\,\,=\,\,$ $\dfrac{-1}{2} \times (-2) \times u^{-\Large \frac{1}{2}}-\dfrac{1}{2} \times c_1$

$\,\,\,=\,$ $\dfrac{(-1) \times (-2)}{2} \times u^{-\Large \frac{1}{2}}-\dfrac{1 \times c_1}{2}$

$\,\,\,=\,\,$ $\dfrac{2}{2} \times u^{-\Large \frac{1}{2}}-\dfrac{1 \times c_1}{2}$

$\,\,\,=\,\,$ $\dfrac{\cancel{2}}{\cancel{2}} \times u^{-\Large \frac{1}{2}}-\dfrac{1 \times c_1}{2}$

$\,\,\,=\,\,$ $1 \times u^{-\Large \frac{1}{2}}-\dfrac{1 \times c_1}{2}$

$\,\,\,=\,\,$ $u^{-\Large \frac{1}{2}}-\dfrac{1 \times c_1}{2}$

$\,\,\,=\,\,$ $\dfrac{1}{u^{\Large \frac{1}{2}}}-\dfrac{1 \times c_1}{2}$

$\,\,\,=\,\,$ $\dfrac{1}{\sqrt{u}}-\dfrac{1 \times c_1}{2}$

The second term in the expression is a constant. Hence, it is simply denoted by a constant $c$.

$\,\,\,=\,\,$ $\dfrac{1}{\sqrt{u}}+c$

The function in this indefinite integral problem is given in terms of $x$ but we obtained the solution in terms of $u$. So, the value of $u$ should be converted back in terms of $x$.

If $ux \,=\, 2+3x$, then $u \,=\, \dfrac{2+3x}{x}$

$\,\,\,=\,\,$ $\dfrac{1}{\sqrt{\dfrac{2+3x}{x}}}+c$

$\,\,\,\therefore\,\,\,\,\,\,$ $\displaystyle \int{\dfrac{1}{x^{\frac{1}{2}}(2+3x)^{\frac{3}{2}}}\,}dx$ $\,=\,$ $\sqrt{\dfrac{x}{2+3x}}+c$

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