Given that $u=x+1$, which integral is equivalent to $\int \frac{x^2}{\sqrt{x+1}} d x$ ?
a $\int(\sqrt{u^3}+\frac{1}{\sqrt{u}}) d u$
b $\int(\sqrt{u^3}-2 \sqrt{u}+\frac{1}{\sqrt{u}}) d u$
c $\int(\sqrt{u^3}-2 \sqrt{u}) d u$
d $\int(\sqrt{u^3}-2 \sqrt{u^2}+\frac{1}{u}) d u$
e $\int(\frac{1}{\sqrt{u}}-2 \sqrt{u}) d u$
Answer (1)
Express x in terms of u : x = u − 1 .
Substitute x = u − 1 into the integral: ∫ u ( u − 1 ) 2 d u .
Expand and simplify the integral: ∫ ( u u 2 − 2 u + 1 ) d u = ∫ ( u 3/2 − 2 u 1/2 + u − 1/2 ) d u .
Rewrite in radical form: ∫ ( u 3 − 2 u + u 1 ) d u . The answer is ∫ ( u 3 − 2 u + u 1 ) d u .
Explanation
Problem Analysis We are given the integral ∫ x + 1 x 2 d x and the substitution u = x + 1 . Our goal is to rewrite the integral in terms of u .
Express x in terms of u and substitute First, we need to express x in terms of u . Since u = x + 1 , we have x = u − 1 . Now we substitute this into the integral: ∫ u ( u − 1 ) 2 d u
Expand the numerator Next, we expand the numerator: ∫ u u 2 − 2 u + 1 d u
Divide by sqrt(u) Now, we divide each term in the numerator by u : ∫ ( u u 2 − u 2 u + u 1 ) d u
Simplify the expression We simplify the expression by using exponent rules. Recall that u = u 1/2 . Thus, we have: ∫ ( u 2 − 1/2 − 2 u 1 − 1/2 + u − 1/2 ) d u = ∫ ( u 3/2 − 2 u 1/2 + u − 1/2 ) d u
Rewrite using radical notation Finally, we rewrite the expression using radical notation: ∫ ( u 3 − 2 u + u 1 ) d u
Final Answer Comparing this result with the given options, we see that it matches option b. Therefore, the equivalent integral is ∫ ( u 3 − 2 u + u 1 ) d u .
Examples
Imagine you are calculating the flow rate of a fluid through a pipe where the velocity profile depends on the position within the pipe. If the velocity is given by a function involving a square root, such as v ( x ) = x + 1 x 2 , and you need to find the total flow, you would integrate this function over the cross-sectional area. Using a substitution like u = x + 1 can simplify the integral, making it easier to compute the total flow rate. This technique is also applicable in various physics and engineering problems involving complex functions under square roots.
