Question

$\int_{}^{}\frac{dx}{(x^{2} + 1)(x^{2} + 4)}$equals:

• A. $\frac{1}{3}$tan^–1 x +$\frac{1}{6}$tan^–1 x + c
• B. $\frac{1}{3}$tan^–1 x –$\frac{1}{6}$tan^–1$\frac{x}{2}$+ c
• C. tan^–1 x –tan^–1$\frac{x}{2}$+ c
• D. 3 tan^–1 x –$\frac{3}{2}$tan^–1$\frac{x}{2}$+ c

Answer 1

Answer: (B)

$\frac{1}{3}$tan^–1 x –$\frac{1}{6}$tan^–1$\frac{x}{2}$+ c

Answer 2

= $\frac { 1 } { 3 }$ $\int_{}^{}\frac{(x^{2} + 4) - (x^{2} + 1)}{(x^{2} + 1)(x^{2} + 4)}$dx

= $\frac { 1 } { 3 }$ $\int_{}^{}\frac{dx}{x^{2} + 4}$

= $\frac{1}{3}$tan^–1 x – $\frac{1}{6}$tan^–1 $\frac{x}{2}$+ c