Question

$\int_{}^{}\frac{x + \sqrt[3]{x^{2}} + \sqrt[6]{x}}{x(1 + \sqrt[3]{x})}$dx equals:

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

Answer 1

Answer: (A)

$\frac{3}{2}$x^2/3 + 6 tan^–1$\sqrt[6]{x}$+ c

Answer 2

Let t⁶ = x

then 6t⁵ dt = dx

I = = 6 $\int_{}^{}\frac{t^{5} + t^{3} + 1}{(1 + t^{2})}$dt

= 6 $\int_{}^{}\frac{t^{3}(t^{2} + 1) + 1}{(1 + t^{2})}$dt

= 6 $\int \mathrm { t } ^ { 3 } \mathrm { dt } + 6$ $\int_{}^{}\frac{1}{1 + t^{2}}$dt

= + 6 tan^–1 t = $\frac{3}{2}$t⁴ + 6 tan^–1 t + c

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