Question

$\int \frac{\sqrt{x}}{\sqrt{x^3+4}}dx$ is equal to

• A. $\frac{2}{3}\ln \left(\frac{2}{\sqrt{x^3}-\sqrt{x^3-4}}\right)+c$
• B. $\frac{2}{3}\ln \left(\frac{2}{\sqrt{x^3}+\sqrt{x^3-4}}\right)+c$
• C. $-\frac{2}{3}\ln \left(\frac{2}{\sqrt{x^3}-\sqrt{x^3-4}}\right)+c$
• D. None of these

Answer 1

Answer: (A)

$\frac{2}{3}\ln \left(\frac{2}{\sqrt{x^3}-\sqrt{x^3-4}}\right)+c$

Given: $\int \frac{\sqrt{x}}{\sqrt{x^3+4}}dx$

Substitute $x^{\frac{3}{2}}=2\tan \theta \Rightarrow \frac{3}{2}{x}^{\frac{1}{2}}dx=2{\sec }^2\theta d\theta$

$\therefore I=\int \frac{\frac{4}{3}{\sec }^2\theta d\theta }{\sqrt{4{\tan }^2\theta +4}}=\frac{2}{3}\int \sec \theta d\theta$

$=\frac{2}{3}\ln (\sec \theta +\tan \theta )+c=\frac{2}{3}\ln (\sqrt{\frac{x^3-4}{4}}+\frac{x^{\frac{3}{2}}}{2})+c$

$=\frac{2}{3}\ln \left(\frac{\sqrt{x^3}+\sqrt{x^3-4}}{2}\right)+c=\frac{2}{3}\ln \left(\frac{x^3-(x^3-4)}{2(\sqrt{x^3}+\sqrt{x^3-4})}\right)+c$

$=\frac{2}{3}\ln \left(\frac{2}{\sqrt{x^3}-\sqrt{x^3-4}}\right)+c$