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What's the definite integral: $$\int_0^{1} \frac{1}{(2x+1)^3} dx$$

I get the answer $\;\;-\dfrac{1}{2(2x+1)^4}$

When I solve I get $(1/162) - (1/2) = -40/81$ , which gives me a negative answer.

What did I do wrong ?

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This is not a primitive. –  1015 Feb 9 '13 at 20:56
    
Evaluating $I=\int (2x+1)^{-3} dx$ should give you something in terms of $(2x+1)^{-2} $, not $(2x+1)^{-4}$. You seem to have differentiated instead of integrated. –  Ganesh Feb 9 '13 at 20:57
    
Thanks for all the helps guy. but i dont understand how u guys change from x=0 to x=1, to x=1 to x=3 –  user61618 Feb 9 '13 at 21:08
    
What's this property or formula comes from ? I don't think I've learned this from my teacher yet. Could you guys show me a website link regarding to this ? –  user61618 Feb 9 '13 at 21:17
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4 Answers

up vote 1 down vote accepted

Hint: let $u = 2x + 1$. Then $du = 2\,dx \implies \,dx = \dfrac 12 \,du$

Your bounds of integration then change: when $x = 0, u = 1$, and when $x = 1, u = 3$.

$$\int_0^{1} \frac{1}{(2x+1)^3} dx = \frac 12 \int_1^3 u^{-3} \,du$$

$$ \frac 12 \int_1^3 u^{-3} \,du = \frac12\cdot -\frac{1}{2} u^{-2}\Big|_1^3 = \dfrac{-1}{4u^2}\Big|_1^3 = -\frac 14 \left(\frac{1}{9} - 1\right) = \frac 29$$

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Thanks for all the helps guy. but i dont understand how u guys change from x=0 to x=1, to x=1 to x=3 –  user61618 Feb 9 '13 at 21:14
    
Is this clear why we need to change the bound of integration? Since we are substituting $u = 2x + 1,$ we need to find the value of $u$ when $x = 0$ (lower bound), and the value of $u$ when $x = 1$.} When $x = 0 \implies u = 2\cdot 0 + 1 = 1$. $x = 1 \implies u = 2\cdot 1 + 1 = 3.$ Since we are substituting to integrate in terms of $u$, we need to use matching parameters for integration –  amWhy Feb 9 '13 at 21:14
    
What's this property or formula comes from ? I don't think I've learned this from my teacher yet. Could you guys show me a website link regarding to this ? –  user61618 Feb 9 '13 at 21:20
    
It comes from the fact that we are substituting $u = 2x + 1$: we use this equation to find the value of $u$ at x = 0, and to find the value of $u$ when x = 1 (the bound of integration are given in terms of x: from 0 to 1, when substituting, we need to get the corresponding bounds in terms of $u$. –  amWhy Feb 9 '13 at 21:24
    
See integration by substitution –  amWhy Feb 9 '13 at 21:32
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First of all, the antiderivative of $(2x+1)^{-3}$ is $-\frac14(2x+1)^{-2}$.

Also, there is a sign error when you plug in the lower limit $x=0$. (You have one minus sign from the antiderivative $F$ and another from the difference $F(1)-F(0)$.

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Consider using $u$-substitution with $u=2x+1$ so that $du=2dx$. Substitution yields $$\frac{1}{2}\int_1^3u^{-3}du=-\frac{1}{4}u^{-2}\bigg|_{u=1}^{u=3}=\frac{-1}{4}\left(\frac{1}{9}-1\right)=2/9.$$

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Thanks for all the helps guy. but i dont understand how u guys change from x=0 to x=1, to x=1 to x=3 –  user61618 Feb 9 '13 at 21:12
    
@user61618: Since we set $u=2x+1$, we plug in the lower bound ($x=0$) to get $u=2(0)+1=1$ and the upper bound ($x=1$) to get $u=2(1)+1=3$. –  Clayton Feb 9 '13 at 21:15
    
Will someone explain the downvote, please? –  Clayton Feb 9 '13 at 21:19
    
What's this property or formula comes from ? I don't think I've learned this from my teacher yet. Could you guys show me a website link regarding to this ? –  user61618 Feb 9 '13 at 21:21
    
See this page on integration by substitution for your questions. It's usually one of the earliest techniques taught for integration. –  Clayton Feb 9 '13 at 21:22
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$$\int_0^{1} \frac{1}{(2x+1)^3} dx$$

Let $2x+1=y$, $2dx=dy$ and \begin{align} \int_0^{1}\rightarrow \int_1^{3}\\ \int_0^{1} \frac{1}{(2x+1)^3} dx&=\int_1^{3} \frac{1}{2y^3} dy\\ &=\left(\frac{-1}{4y^2}\right)_1^3\\ &=\vdots \end{align} You messed up the last step.

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@julien. Must be careful during copy-pasting. Thanks –  Nitish Feb 9 '13 at 21:02
    
Thanks for all the helps guy. but i dont understand how u guys change from x=0 to x=1, to x=1 to x=3 –  user61618 Feb 9 '13 at 21:13
    
@user61618, $2x+1=y$ so, when $x=0, y=2\times 0 + 1 = 1$, and similarly, when $x=1,y=3$. –  Nitish Feb 9 '13 at 21:16
    
What's this property or formula comes from ? I don't think I've learned this from my teacher yet. Could you guys show me a website link regarding to this ? –  user61618 Feb 9 '13 at 21:21
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