Evaluate

∫ [(5+x)^2]/√x

x^-.5 ( x^2 + 10 x + 25) dx

x^(3/2)dx + 10x^(1/2)dx + 25x^(-1/2)dx

(2/5)x^(5/2) + 10 ... etc

if u don't mind could you complete it. please im confused and i have an exam tommorow. please Damon !

+ 10(2/3) x^(3/2) + 25(2) x^(1/2)

To evaluate the integral ∫ [(5+x)^2]/√x, we can use a technique called u-substitution.

Let's start by letting u = 5 + x. Taking the derivative of u with respect to x gives us du = dx.

Next, we need to express the original integral in terms of u. To do this, we need to solve for x in terms of u. Rearranging the equation u = 5 + x, we get x = u - 5.

Now, substitute u = 5 + x and dx = du into the original integral:
∫ [(5+x)^2]/√x dx becomes ∫ [u^2]/√(u-5) du.

We can simplify the expression to ∫ (u^2)/(√u - √5) du.

Now, let's focus on simplifying the expression inside the integral. Notice that the denominator has a difference of two square roots. We can multiply the denominator by its conjugate to eliminate the square root:

(√u - √5)(√u + √5)

Using the difference of squares formula, (√a - √b)(√a + √b) = a - b, we get:

u - 5

Now our integral becomes ∫ (u^2)/(u - 5) du.

We can now divide the numerator by the denominator using long division or synthetic division to simplify the integrand:
u^2 ÷ (u - 5) = (u + 5) + 25/(u - 5)

Our integral becomes ∫ [(u + 5) + 25/(u - 5)] du.

Now we can split the integral into two separate integrals:
∫ (u + 5) du + ∫ 25/(u - 5) du.

Integrating (u + 5) with respect to u gives us (1/2)u^2 + 5u.

For the second integral, we can use the substitution method. Let v = u - 5, which means du = dv.

Now the integral becomes ∫ 25/v dv, which is a simple integral to solve:
∫ 25/v dv = 25 ln|v| + C.

Substituting the original variable back in, we have 25 ln|u - 5| + C.

Finally, we have our evaluated integral:
∫ [(5+x)^2]/√x dx = (1/2)(5 + x)^2 + 5(5 + x) + 25 ln|5 + x - 5| + C.

Simplifying further, we get:
(1/2)(5 + x)^2 + 5(5 + x) + 25 ln|x| + C.

And that is our final answer. The integral evaluates to (1/2)(5 + x)^2 + 5(5 + x) + 25 ln|x| + C.