Integration by substitution thread 2?

**homeylova223** · August 18th 2011, 09:04 PM

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.

**Prove It** · August 18th 2011, 09:09 PM

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

**Soroban** · August 19th 2011, 04:07 AM

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

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

Thread: Integration by substitution thread 2?

LinkBack

Thread Tools

Display

Integration by substitution thread 2?

Re: Integration by substitution thread 2?

Re: Integration by substitution thread 2?

Similar Math Help Forum Discussions

The natural logarithmic function and integration thread 2?

Integration by substitution thread 3?

Integration by substitution

integration by substitution

integration by substitution

Search Tags