1. ## Integration by substitution thread 2?

I am wondering if I did this correctly.

Integrate

x square root(2x+1)

u=2x+1 x=(1/2)(u-1)

(1/2)(u-1) square root(u)

(1/2) (u)^(3/2)-(u)^(1/2)

(1/2) (2/5)(u)^(5/2)-(2/3)(u)^(3/2)

(2/10)u^(5/2)-(2/6)(u)^(3/2)+c

Any help would be appreciated on this difficult topic.

2. ## Re: Integration by substitution thread 2?

Not quite. You are trying to integrate $\displaystyle \int{x\sqrt{2x + 1}\,dx}$.

If you let $\displaystyle u = 2x + 1$ then $\displaystyle x = \frac{1}{2}(u - 1)$ AND $\displaystyle du = 2\,dx$.

So the integral becomes

\displaystyle \begin{align*}\int{x\sqrt{2x + 1}\,dx} &= \frac{1}{2}\int{x\sqrt{2x + 1}\cdot 2\,dx} \\ &= \frac{1}{2}\int{\frac{1}{2}(u - 1)\sqrt{u}\,du} \\ &= \frac{1}{4}\int{(u - 1)u^{\frac{1}{2}}\,du} \\ &= \frac{1}{4}\int{u^{\frac{3}{2}} - u^{\frac{1}{2}}\,du} \end{align*}

Go from here

3. ## Re: Integration by substitution thread 2?

Hello, homeylova223!

$\int x\sqrt{2x+1}\,dx$

Try the method I suggested in your previous post.

Let $u \,=\,\sqrt{2x+1} \quad\Rightarrow\quad x \,=\,\frac{u^2-1}{2} \quad\Rightarrow\quad dx \,=\,u\,du$

Substitute: . $\int \frac{u^2-1}{2}\cdot u\cdot u\,du \;=\;\tfrac{1}{2}{\int(u^4 - u^2)\,du$

. . . . $=\;\tfrac{1}{2}\left(\tfrac{1}{5}u^5 - \tfrac{1}{3}u^3\right) + C \;=\; \frac{u^3}{30}(3u^2-5) + C$

Back-substitute: . $\frac{(\sqrt{2x+1})^3}{30}\bigg[3(2x+1)-5\bigg] + C$

. . $=\;\tfrac{1}{30}(2x+1)^{\frac{3}{2}}(6x-2) + C \;=\;\tfrac{1}{15}(2x+1)^{\frac{3}{2}}(3x-1) + C$