Find the most general antiderivative.
-$$\int \frac { x \sqrt { x } + \sqrt { x } } { x ^ { 2 } } d x$$
A) $- \frac { \sqrt { x } } { 2 } - \frac { 3 \sqrt { x } } { 2 } + C$
B) $2 \sqrt { x } - \frac { 2 } { \sqrt { x } } + C$
C) $\mathrm { C }$
D) $\frac { 2 } { \sqrt { x } } - 2 \sqrt { x } + C$

Question

Quizplus · Accepted Answer

We can rewrite the given expression as:
$$\int \frac{x\sqrt{x}+\sqrt{x}}{x^2}dx = \int \frac{\sqrt{x}(x+1)}{x^2}dx = \int \frac{x^{-\frac{3}{2}}(x+1)}{(x^{-\frac{1}{2}})}dx$$
Now, using the power rule and simplifying:
$$\int \frac{x^{-\frac{3}{2}}(x+1)}{(x^{-\frac{1}{2}})}dx = \int \frac{x^{-2}+x^{-\frac{3}{2}}}{dx} = -2\sqrt{x}-2x^{-\frac{1}{2}}+C$$
Simplifying further gives us the desired antiderivative of:
$$2\sqrt{x}-\frac{2}{\sqrt{x}}+C$$

Find the Most General Antiderivative ∫xx+xx2dx\int \frac { x \sqrt { x } + \sqrt { x } } { x ^ { 2 } } d x∫x2xx​+x​​dx

Find the Most General Antiderivative $\int \frac { x \sqrt { x } + \sqrt { x } } { x ^ { 2 } } d x$