definite integral question?

Question

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 ?

Answer 1

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 $x = 1, u = 3$.

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

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

Answer 2

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 $F(1)-F(0)$.

Answer 3

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

Answer 4

$$\int_0^{1} \frac{1}{(2x+1)^3} dx$$

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.