# Math Help - integrate

1. ## integrate

integrate $\frac{4x}{\sqrt{2x+1}}$

2. $\int{\frac{4x}{\sqrt{2x+1}}\,dx}$.

Let $u = 2x + 1$ so that $x = \frac{u-1}{2}$ and $du = 2\,dx$.

The integral becomes:

$\int{\frac{4x}{\sqrt{2x+1}}\,dx} = \int{\frac{2x}{\sqrt{2x + 1}}\,2\,dx}$

$= \int{\frac{u - 1}{\sqrt{u}}\,du}$

$= \int{u^{\frac{1}{2}} - u^{-\frac{1}{2}}\,du}$

$= \frac{u^{\frac{3}{2}}}{\frac{3}{2}} - \frac{u^{\frac{1}{2}}}{\frac{1}{2}} + C$

$= \frac{2\sqrt{(2x + 1)^3}}{3} - 2\sqrt{2x+1} + C$

$= \frac{2\sqrt{(2x + 1)^3} - 6\sqrt{2x + 1}}{3} + C$

$= \frac{2(2x + 1)\sqrt{2x + 1} - 6\sqrt{2x + 1}}{3} + C$

$= \frac{(4x + 2)\sqrt{2x + 1} - 6\sqrt{2x + 1}}{3} + C$

$= \frac{(4x - 4)\sqrt{2x + 1}}{3} + C$

$= \frac{4(x - 1)\sqrt{2x + 1}}{3} + C$.

3. Hello, Punch!

Another method . . .

Integrate: . $\displaystyle{\int \frac{4x}{\sqrt{2x+1}}}\,dx$

We have: . $\displaystyle{4\int\frac{x}{\sqrt{2x+1}} \,dx
}$

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

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

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

Back-substitute: . $\frac{2}{3}\sqrt{2x+1}\,(2x+1 - 3) + C$

. . . . . . . . . . $=\; \frac{2}{3}\sqrt{2x+1}\,(2x - 2) + C$

. . . . . . . . . . $=\; \frac{4}{3}\sqrt{2x+1}\,(x-1) + C$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

When given a linear expression under a radical,
. . let $u \,=\text{ the entire radical}$

Given: .an integral with $\sqrt[n]{ax+b}$

$\text{Let: }\:u \:=\:\sqrt[n]{ax+b} \quad\Rightarrow\quad x \:=\:\dfrac{u^n - b}{a} \quad\Rightarrow\quad dx \:=\:\frac{n}{a}u^{n-1}du$

. . and the new integral will have no radicals (usually).

4. And another solution...

By substitution: x=(1/2)cos(2t). dx=-sin(2t), sqrt(2x+1)=sqrt(cos(2t)+1)=sqrt(2cos^2(t))=sqrt(2) *cos(t)

Ok... maybe it's a little bit complicated...