What is the antiderivative of this?
What is the antiderivative of x/((16+x^2)^(1/2))?
(x / the square root of (16 + x^2))?
What method did you use? Substitution, etc..?
Thanks!
Let 16 + x^2 = u
2x dx = du
Your integral becomes the integral of
(1/2) u^-1/2 du
which is
u^1/2 = sqrt(16 + x^2)
Find the most general antiderivative of the function. (Check your answer by differentiation. Use C for the constant of the antiderivative.)
f(x) = (x + 3)(5x − 1)
To find the antiderivative of the function x/((16+x^2)^(1/2)), we can use a method called substitution.
First, let's choose a new variable, u, and set it equal to the expression inside the square root, 16 + x^2. So, u = 16 + x^2.
Next, we need to find du/dx, which is the derivative of u with respect to x. Differentiating both sides of the equation u = 16 + x^2, we get du/dx = 2x.
Now, let's solve for dx. Rearranging the equation, we have dx = (du/2x).
Substituting the values dx and u into the original function, we get:
∫ (x/((16+x^2)^(1/2))) dx = ∫ (x/((u)^(1/2))) (du/2x).
The x in the numerator and the x in the denominator cancel out, leaving us with:
∫ (1/((u)^(1/2))) (du/2).
Simplifying this expression, we have:
(1/2) ∫ (1/((u)^(1/2))) du.
Now, we can rewrite the integral in terms of the new variable u:
(1/2) ∫ (u^(-1/2)) du.
To find the antiderivative of u^(-1/2), we can add 1 to the exponent and divide by the new exponent:
(1/2) ∫ (u^(1/2)) / (1/2) du = (1/2) ∫ u^(1/2) du.
Integrating u^(1/2) with respect to u, we get:
(1/2) * (2/3) * u^(3/2) + C.
Finally, substituting back u = 16 + x^2, we have:
(1/2) * (2/3) * (16 + x^2)^(3/2) + C, where C is the constant of integration.
In conclusion, the antiderivative of x/((16+x^2)^(1/2)) is (1/3) * (16 + x^2)^(3/2) + C, where C is the constant of integration.