Question

Integrate the function $\sqrt {{x^2} + 3x}$

Answer 1

According to the question we have to integrate the given function which is $\sqrt {{x^2} + 3x}$. So, first of all we have to make the terms of the given function a whole square for which we have to add and subtract the terms that are missing in completing the square.
Now, we have to make the term inside the form of whole square and then to solve the expression above we have to use the formula as mentioned below:

Formula used:
$\Rightarrow {(a + b)^2} = {a^2} + {b^2} + 2ab................(A)$
Now, applying the formula (A) just above, now we have to find the integration of the obtained expression and to find the integration we have to use the formula as mentioned below:

$\Rightarrow \int {\sqrt {{x^2} - {a^2}} dx = \dfrac{x}{2}} \sqrt {{x^2} - {a^2}} - \dfrac{{{a^2}}}{2}\log \left| {x + \sqrt {{x^2} - {a^2}} } \right| + C$………………..(B)
Hence, with the help of the formula (B) we can determine the integration of the given function.

Complete step by step answer:
Step 1: First of all we have to let the given function as I mentioned below to determine the integration.
$I = \int {\sqrt {{x^2} + 3x} } dx$
Step 2: Now, we have to make the terms of the given function a whole square for which we have to add and subtract the terms that are missing in completing the square as mentioned in the solution hint so we have to $\dfrac{9}{4}$ add and subtract in the given function. Hence,
$I = \int {\sqrt {{x^2} + 3x + \dfrac{9}{4} - \dfrac{9}{4}} } dx$
Now, we have to write $\dfrac{9}{4}$ in the form of whole square,

$$I = \int {\sqrt {{x^2} + 3x + {{\left( {\dfrac{3}{4}} \right)}^2} - {{\left( {\dfrac{3}{4}} \right)}^2}} } dx \\\ I = \int {\sqrt {{{\left( {x + \dfrac{3}{2}} \right)}^2} - {{\left( {\dfrac{3}{4}} \right)}^2}} dx} ...............(1) \\\$$

Step 3: Now, to solve the integration as obtained in the solution step 2 we have to use the formula (B) as mentioned in the solution hint. Hence,
$\Rightarrow \dfrac{{\left( {x + \dfrac{3}{2}} \right)}}{2}\sqrt {{x^2} + 3x} - \dfrac{{\dfrac{9}{4}}}{2}\log \left| {\left( {x + \dfrac{3}{2}} \right) + \sqrt {{x^2} + 3x} } \right| + C$
Where C is the constant and now we have to solve the integrated function as obtained just above, Hence,
$\Rightarrow \dfrac{{2x + 3}}{4}\sqrt {{x^2} + 3x} - \dfrac{9}{8}\log \left| {\left( {x + \dfrac{3}{2}} \right) + \sqrt {{x^2} + 3x} } \right| + C$
Hence, with the help of the formula (A) and (B) we have determine the integration of the given function $\sqrt {{x^2} + 3x}$ is $\dfrac{{2x + 3}}{4}\sqrt {{x^2} + 3x} - \dfrac{9}{8}\log \left| {\left( {x + \dfrac{3}{2}} \right) + \sqrt {{x^2} + 3x} } \right| + C$.

Note: To obtain the integration of the given function it is necessary that we have to determine the terms of the function in the form of a whole square which can be done by adding and subtracting the terms which are missing to form a complete square for the given fraction.