Question

If I = $\int_{}^{}\frac{\sqrt{x^{2} + 1}}{x^{4}}$dx, then I equals

• A. – $\frac { 1 } { 3 }$ $\frac{(x^{2} + 1)^{3/2}}{x^{3}}$+ C
• B. x³ (x² + 1)^–1/2 + C
• C. $\frac{\sqrt{x^{2} + 1}}{x^{2}}$+ C
• D. –$\frac { 1 } { 3 }$ $\frac{(x^{2} + 1)^{3/2}}{x^{2}}$+ C

Answer 1

Answer: (A)

– \frac { 1 } { 3 }$$\frac{(x^{2} + 1)^{3/2}}{x^{3}}+ C

Answer 2

Put x = tan q so that $\sqrt{x^{2} + 1}$= sec q , dx = sec² q dq

\ I = $\int_{}^{}\frac{\sec\theta\sec^{2}\theta}{\tan^{4}\theta}$dq = $\int_{}^{}\frac{\cos\theta}{\sin^{4}\theta}$dq

= – $\frac { 1 } { 3 }$ $\frac{1}{\sin^{3}\theta}$+ C

= – $\frac { 1 } { 3 }$ $\frac{(x^{2} + 1)^{3/2}}{x^{3}}$+ C