integral of inverse root(8x+x^2)
To find the integral of the function f(x) = 1/√(8x + x²), we first need to simplify the expression under the square root.
Notice that the expression 8x + x² can be rewritten as (x² + 8x), which suggests completing the square to make the integration easier.
To complete the square, we divide the coefficient of x (which is 8) by 2 and then square the result: (8/2)² = 16.
Next, we add and subtract this value inside the square root:
f(x) = 1/√(x² + 8x + 16 - 16)
= 1/√[(x + 4)² - 16]
Now, we can factor out the square of the binomial using the identity (a - b)² = a² - 2ab + b²:
f(x) = 1/√[(x + 4)² - 4²]
= 1/√[(x + 4 - 4)(x + 4 + 4)]
= 1/√[(x)²]
Simplifying further:
f(x) = 1/|x|
Now, we can integrate f(x) = 1/|x| by splitting it into two separate cases, depending on whether x is positive or negative:
For x > 0:
∫(1/x) dx = ln|x| + C₁
For x < 0:
∫(1/x) dx = ln(-x) + C₂
where C₁ and C₂ are constants of integration.
Thus, the integral of the original function f(x) = 1/√(8x + x²) is given by:
∫[f(x)] dx = ln|x| + C₁ (for x > 0)
ln(-x) + C₂ (for x < 0)
Note that the absolute value notation is used to account for the two different cases.