Question

Evaluate: $\int \frac{3x^2}{4+5x^6}dx$

• A. $\frac{1}{\sqrt{\left(5\right)}}{\tan }^{-1}\frac{\sqrt{5}{x}^3}{2}.$
• B. $\frac{1}{2\sqrt{\left(5\right)}}{\tan }^{-1}\frac{\sqrt{3}{x}^3}{2}.$
• C. $\frac{1}{2\sqrt{\left(5\right)}}{\tan }^{-1}\frac{\sqrt{5}{x}^5}{2}.$
• D. $\frac{1}{2\sqrt{\left(5\right)}}{\tan }^{-1}\frac{\sqrt{5}{x}^3}{2}.$

Answer 1

Answer: (D)

$\frac{1}{2\sqrt{\left(5\right)}}{\tan }^{-1}\frac{\sqrt{5}{x}^3}{2}.$

Let $I=\int \frac{3x^2}{4+5x^6}dx$

Let $u=x^3$

$\frac{du}{dx}=3x^2$

$du=3x^{2}dx$

$I=\int \frac{du}{4+5u^2}$

Now, apply integral substitution,

$u=\frac{2}{\sqrt{5}}v$

$I=\int \frac{dv}{2\sqrt{5}(v^2+1)}dv$

$I=\frac{1}{2\sqrt{5}}\int \frac{1}{v^2+1}dv$

$I=\frac{1}{2\sqrt{5}}{\tan }^{-1}v$

$I=\frac{1}{2\sqrt{5}}{\tan }^{-1}\left(\frac{\sqrt{5}}{2}u\right)$ [ Re-substituting value of $v$ ]

$I=\frac{1}{2\sqrt{5}}{\tan }^{-1}\left(\frac{\sqrt{5}}{2}{x}^3\right)$ [ Re-substituting value of $u$ ]