Question

What would be the anti-derivative of (x-7) / (x)^0.5 ?

Answer 1

Answer:

$\tfrac23x^{3/2} - 14x^{1/2} + C$

Step-by-step explanation:

$\displaystyle \int \frac{x-7}{x^{1/2}}\,dx\\= \int\bigl(x^{1/2} - 7x^{-1/2}\bigr)\,dx\\=\tfrac23x^{3/2} - 7\bigl(\tfrac21x^{1/2}\bigr) + C\\=\tfrac23x^{3/2} - 14x^{1/2} + C$