Question

$\int_{}^{}{\frac{\mathbf{(x + 1}\mathbf{)}^{\mathbf{2}}}{\mathbf{x(}\mathbf{x}^{\mathbf{2}}\mathbf{+ 1)}}\mathbf{dx}}$is equal to

• A. $\sec x - x + c$
• B. $\int_{}^{}{e^{\log(\sin x)}dx =}$
• C. $\sin x + c$
• D. $- \cos x + c$

Answer 1

Answer: (B)

$\int_{}^{}{e^{\log(\sin x)}dx =}$

Answer 2

$\int_{}^{}{\frac{(x + 1)^{2}}{x(x^{2} + 1)}dx =}\int_{}^{}{\frac{x^{2} + 1 + 2x}{x((x^{2} + 1)}dx} = \int_{}^{}{\frac{x^{2} + 1}{x(x^{2} + 1)}dx + 2\int_{}^{}{\frac{x}{x(x^{2} + 1)}dx}} = \int_{}^{}{\frac{dx}{x} + 2}\int_{}^{}{\frac{dx}{x^{2} + 1} = \log_{e}x + 2\tan^{- 1}x} + c$