Question

$\int \frac{x+1}{2x^{\frac{3}{2}}}dx$ is equal to (where C is constant of integration)

• A. $x^{\frac{1}{2}}-x^{\frac{-1}{2}}+C$
• B. $\frac{x^{\frac{3}{2}}-x^{\frac{1}{2}}}{x}+C$
• C. $\frac{x^{\frac{3}{2}}+\sqrt{x}(\sqrt{x}-1)}{x}+C$
• D. $\frac{x^2-1}{x^{\frac{3}{2}}+x^{\frac{1}{2}}}+C$

Answer 1

Answer: (A), (B), (C), (D)

$\frac{x^2-1}{x^{\frac{3}{2}}+x^{\frac{1}{2}}}+C$

$I=\frac{1}{2}\int (x^{\frac{-1}{2}}+x^{\frac{-3}{2}})dx=\frac{1}{2}[2\sqrt{x}-\frac{2}{\sqrt{x}}]+CI=\sqrt{x}-\frac{1}{\sqrt{x}}+C=\frac{x-1}{\sqrt{x}}+C=\frac{x^{\frac{3}{2}}-x^{\frac{1}{2}}}{x}+(C+1)$
(c): $\frac{x^{\frac{3}{2}}+x-x^{\frac{1}{2}}}{x}+C=\frac{x^{\frac{3}{2}}-x^{\frac{1}{2}}}{x}+(C+1)$
(d): $\frac{(x^2-1)}{x^{\frac{3}{2}}+x^{\frac{1}{2}}}=\frac{(x-1)(x+1)}{x^{\frac{1}{2}}(1+x)}=x^{\frac{1}{2}}-x^{-\frac{1}{2}}+C$