Rooted Integrand

Calculus Level 4

$\int _{ 0 }^{ 64 }{ \frac { 1 }{ { x }^{ \frac { 1 }{ 2 } }+{ x }^{ \frac { 1 }{ 3 } } } } dx = \ ?$

1 solution

คลุง แจ็ค
Apr 28, 2015

Write a solution. We will start with changing the variables

Let's $x=u^{6}$ then

$\int \frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}dx=\int \frac{6u^{5}}{u^{3}+u^{2}}du$

This integral has the solution to be

$6(\frac{u^{3}}{3}-\frac{u^{2}}{2}+u-ln(u+2))$ from u=0 to 2

Plug this value to this definite integral we have

$16-6\times ln(3)\approx 9.408$

You should clarify why you use the substitution $x = u^6$ . And you could simply the integral further after substitution.

Similar method as yours, sir.

Noel Lo - 6 years, 1 month ago

I use $x=u^{6}$ 'cause it will make the power of changed variable to be an integer. So it'll simplify our calculation.

คลุง แจ็ค - 6 years, 1 month ago

@Utkarsh Bansal hey many of your answers come in decimal , btw enjoy solving , thanks for the problem ..

Rudraksh Sisodia - 4 years, 9 months ago