Easy definite one

Calculus Level 1

∫ 0 1 1 x ( 1 + x ) 2 d x = ? \large \int_0^1 \frac 1{\sqrt x(1+\sqrt x)^2} dx = \ ?


The answer is 1.

José Antonio Fuentes by José Antonio Fuentes , 18, Spain

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3 solutions

Chew-Seong Cheong
Feb 29, 2020

@Alak Bhattacharya's solution in LaTex

I = ∫ 0 1 1 x ( 1 + x ) 2 d x Let u = 1 + x ⟹ d u = d x 2 x = ∫ 1 2 2 u 2 d u = − 2 u ∣ 1 2 = 1 \begin{aligned} I & = \int_0^1 \frac 1{\sqrt x(1+\sqrt x)^2} dx & \small \blue{\text{Let }u = 1 + \sqrt x \implies du = \frac {dx}{2\sqrt x}} \\ & = \int_1^2 \frac 2{u^2} du = - \frac 2u\ \bigg|_1^2 = \boxed 1 \end{aligned}

0 Helpful 0 Interesting 2 Brilliant 0 Confused

I think you have a typo on your limits of integration ( first integral limits after the change of variables should be in terms of u u as well )?

Eric Roberts - 1 year, 3 months ago

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Thanks. I have amended it.

Chew-Seong Cheong - 1 year, 3 months ago

Substitute 1 + x = y 1+\sqrt x=y . Then d x 2 x = d y \dfrac{dx}{2\sqrt x}=dy . At x = 0 , y = 1 x=0,y=1 and at x = 1 , y = 2 x=1,y=2 . Hence the value of the integral is 2 ( 1 − 1 2 ) = 1 2(1-\dfrac{1}{2})=\boxed 1

2 Helpful 0 Interesting 0 Brilliant 0 Confused

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice m8 you are really clever

Juan Manuel García - 1 year, 3 months ago

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