Integral Cake Delivery IIIIII

Calculus Level 4

∫ 1 e x 3 x ( x + x 3 ) d x \large \int_1^e \dfrac{\sqrt[3]{x} }{ x (\sqrt x + \sqrt[3]{x} )} \, dx

If the value of the integral above is equal to a − b ln ⁡ ( c + e d ) + f ln ⁡ g , a- b \ln(c+ \sqrt[d]{e} ) + f \ln g,

where a , b , c , d , f a,b,c,d,f and g g are positive integers with b b and g g minimized, find a + b + c + d + f + g a+b+c+d+f+g .


The answer is 22.

View wiki
O D by O D , 32, Antarctica

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

I = ∫ 1 e x 3 x ( x + x 3 ) d x Divide up and down by x 3 = ∫ 1 e 1 x ( x 6 + 1 ) d x Let u = x 6 ⟹ 6 u 5 d u = d x = ∫ 1 e 6 6 u 5 u 6 ( u + 1 ) d u = 6 ∫ 1 e 6 1 u ( u + 1 ) d u = 6 ∫ 1 e 6 ( 1 u − 1 u + 1 ) d u = 6 [ ln ⁡ u − ln ⁡ ( u + 1 ) ] 1 e 6 = 1 − 6 ln ⁡ ( 1 + e 6 ) + 6 ln ⁡ 2 \begin{aligned} I & = \int_1^e \frac {\sqrt[3]x}{x\left(\sqrt x+\sqrt[3]x\right)} dx & \small \color{#3D99F6} \text{Divide up and down by }\sqrt[3]x \\ & = \int_1^e \frac 1{x\left(\sqrt[6]x+ 1\right)} dx & \small \color{#3D99F6} \text{Let }u = \sqrt[6]x \implies 6u^5 \ du = dx \\ & = \int_1^{\sqrt[6]e} \frac {6u^5}{u^6\left(u+ 1\right)} du \\ & = 6 \int_1^{\sqrt[6]e} \frac 1{u\left(u+ 1\right)} du \\ & = 6 \int_1^{\sqrt[6]e} \left( \frac 1u - \frac 1{u+ 1}\right) du \\ & = 6 \bigg[\ln u - \ln(u+1) \bigg]_1^{\sqrt[6]e} \\ & = 1 - 6\ln \left(1+\sqrt[6]e \right) + 6\ln 2 \end{aligned}

⟹ a + b + c + d + f + g = 1 + 6 + 1 + 6 + 6 + 2 = 22 \implies a+b+c+d+f+g = 1 + 6 + 1 + 6 + 6 + 2 = \boxed{22}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Wow, the writer of this problem is in Antarctica?! (I had the same solution by the way. Nice job, as always!)

James Wilson - 3 years, 6 months ago

Log in to reply

Wonder if he is lonely there.

Chew-Seong Cheong - 3 years, 6 months ago

Log in to reply

I clicked on the profile, but apparently the account doesn't even exist anymore.

James Wilson - 3 years, 6 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...