Answer :
To solve the integral [tex]\(\int \frac{dx}{\sqrt{2ax + x^2}}\)[/tex], let's go through the detailed steps:
### Step 1: Recognize the Structure of the Integrand
The integrand can be written as:
[tex]\[ \frac{1}{\sqrt{2ax + x^2}} \][/tex]
Notice that [tex]\(2ax + x^2\)[/tex] can be factored and simplified further.
### Step 2: Completing the Square
Rewrite the expression [tex]\(2ax + x^2\)[/tex]:
[tex]\[ 2ax + x^2 = x^2 + 2ax \][/tex]
Complete the square for the quadratic expression:
[tex]\[ x^2 + 2ax = (x + a)^2 - a^2 \][/tex]
So, the integral becomes:
[tex]\[ \int \frac{dx}{\sqrt{(x + a)^2 - a^2}} \][/tex]
### Step 3: Substitution to Simplify the Integral
Let [tex]\( u = x + a \)[/tex]. Then, [tex]\( du = dx \)[/tex]. The limits of integration and [tex]\(dx\)[/tex] directly translate into the new variable [tex]\(u\)[/tex]:
[tex]\[ \int \frac{du}{\sqrt{u^2 - a^2}} \][/tex]
This is a standard integral form relating to the hyperbolic functions.
### Step 4: Using the Standard Integral Form
The integral [tex]\(\int \frac{du}{\sqrt{u^2 - a^2}}\)[/tex] matches the standard integral:
[tex]\[ \int \frac{du}{\sqrt{u^2 - a^2}} = \log |u + \sqrt{u^2 - a^2}| + C \][/tex]
Substitute back [tex]\(u = x + a\)[/tex]:
[tex]\[ \log |x + a + \sqrt{(x + a)^2 - a^2}| + C \][/tex]
### Step 5: Simplifying Back into Original Variables
Recall [tex]\((x + a)^2 - a^2 = 2ax + x^2\)[/tex]:
[tex]\[ \log |x + a + \sqrt{2ax + x^2}| + C \][/tex]
### Final Answer
Combining all steps, the final result is:
[tex]\[ \boxed{\log |x + a + \sqrt{2ax + x^2}| + C} \][/tex]
However, considering another interpretation, it could also be expressed using piecewise functions based on computing conditions. If the integral involves a condition where [tex]\(a = 0\)[/tex], a different representation might come into play. Hence, considering conditions of [tex]\(a\)[/tex], the final result could be expressed in a piecewise manner as:
[tex]\[ \boxed{ \begin{cases} \log |2a + 2x + 2\sqrt{2ax + x^2}| & \text{if } a^2 \neq 0 \\ (x + a) \log |a + x| / \sqrt{(a + x)^2} & \text{otherwise} \end{cases} } \][/tex]
### Step 1: Recognize the Structure of the Integrand
The integrand can be written as:
[tex]\[ \frac{1}{\sqrt{2ax + x^2}} \][/tex]
Notice that [tex]\(2ax + x^2\)[/tex] can be factored and simplified further.
### Step 2: Completing the Square
Rewrite the expression [tex]\(2ax + x^2\)[/tex]:
[tex]\[ 2ax + x^2 = x^2 + 2ax \][/tex]
Complete the square for the quadratic expression:
[tex]\[ x^2 + 2ax = (x + a)^2 - a^2 \][/tex]
So, the integral becomes:
[tex]\[ \int \frac{dx}{\sqrt{(x + a)^2 - a^2}} \][/tex]
### Step 3: Substitution to Simplify the Integral
Let [tex]\( u = x + a \)[/tex]. Then, [tex]\( du = dx \)[/tex]. The limits of integration and [tex]\(dx\)[/tex] directly translate into the new variable [tex]\(u\)[/tex]:
[tex]\[ \int \frac{du}{\sqrt{u^2 - a^2}} \][/tex]
This is a standard integral form relating to the hyperbolic functions.
### Step 4: Using the Standard Integral Form
The integral [tex]\(\int \frac{du}{\sqrt{u^2 - a^2}}\)[/tex] matches the standard integral:
[tex]\[ \int \frac{du}{\sqrt{u^2 - a^2}} = \log |u + \sqrt{u^2 - a^2}| + C \][/tex]
Substitute back [tex]\(u = x + a\)[/tex]:
[tex]\[ \log |x + a + \sqrt{(x + a)^2 - a^2}| + C \][/tex]
### Step 5: Simplifying Back into Original Variables
Recall [tex]\((x + a)^2 - a^2 = 2ax + x^2\)[/tex]:
[tex]\[ \log |x + a + \sqrt{2ax + x^2}| + C \][/tex]
### Final Answer
Combining all steps, the final result is:
[tex]\[ \boxed{\log |x + a + \sqrt{2ax + x^2}| + C} \][/tex]
However, considering another interpretation, it could also be expressed using piecewise functions based on computing conditions. If the integral involves a condition where [tex]\(a = 0\)[/tex], a different representation might come into play. Hence, considering conditions of [tex]\(a\)[/tex], the final result could be expressed in a piecewise manner as:
[tex]\[ \boxed{ \begin{cases} \log |2a + 2x + 2\sqrt{2ax + x^2}| & \text{if } a^2 \neq 0 \\ (x + a) \log |a + x| / \sqrt{(a + x)^2} & \text{otherwise} \end{cases} } \][/tex]