How to integrate $\int\frac{2x+1}{\sqrt{x+3}}dx$?

Question

I'm trying to solve this:

$$\int\frac{2x+1}{\sqrt{x+3}}dx$$

I was thinking in use substitution technique but if I take my $u$ as $x+3$, my $du$ isn't what is left.

Answer 1

I think you are on the right track but just didn't complete the thought:

Taking $u=x+3 \implies du= dx$

Note that $2x+1= 2(x+3)-5=2u-5$

$$\int\frac{2x+1}{\sqrt{x+3}}dx= \int\frac{2u-5}{\sqrt{u}}du=2\int u^{1/2}du-5\int u^{-1/2}du$$

What's next?

Answer 2

$$\int\frac{2x+1}{\sqrt{x+3}}dx=\int\frac{2(x+3)-5}{\sqrt{x+3}}d(x+3)=\int\frac{2u-5}{\sqrt u}du$$ $du=dx$ is not always true, but $dx=\frac{dx}{du}du$ is true. Or if you want to get rid of notation $d(x+3)$, just use $$\int\frac{2x+1}{\sqrt{x+3}}dx=\int\frac{2u-5}{\sqrt u}\frac{dx}{du}du=\int\frac{2u-5}{\sqrt u}du$$

Answer 3

Using $2x+1 = 2(x+3) - 5$, your integral becomes: \begin{equation} \int\frac{2x+1}{\sqrt{x+3}}dx = \underbrace{{\displaystyle\int}\sqrt{x+3}\,\mathrm{d}x}_A- \underbrace{{\displaystyle\int}\dfrac{1}{\sqrt{x+3}}\,\mathrm{d}x}_B \end{equation} We know that \begin{align} A &= \dfrac{2\left(x+3\right)^\frac{3}{2}}{3}\\ B &= 2\sqrt{x+3} \end{align}

Answer 4

Note that $u$ substitution with your recommendation, $u=x+3$, yields the integral $$\int \frac{2x+1}{\sqrt{u}} du.$$ Let us try to get this integral entirely in terms of $u$. Notice that $x=u-3$, so the integral will turn out to be $$\int \frac{2(u-3)+1}{\sqrt{u}}du.$$ Now expanding the numerator and then dividing each term in the numerator by $\sqrt{u}$ will let you use the reverse power rule to get a solution.

Answer 5

Here's a more general method.

Consider the integral $$I=\int \frac{a_1x+a_2}{\sqrt{a_3x+a_4}}dx$$ We may preform the substitution $x=\frac{u-a_4}{a_3}$ $$I=\frac1{a_3}\int \frac{\frac{a_1}{a_3}u-\frac{a_4}{a_3}+a_2}{u^{1/2}}du$$ $$I=\frac1{a_3^2}\int\frac{u}{u^{1/2}}du+\frac1{a_3}\bigg(a_2-\frac{a_4}{a_3}\bigg)\int\frac{du}{u^{1/2}}$$ $$I=\frac{2u^{3/2}}{3a_3^2}+\frac{2u^{1/2}}{a_3}\bigg(a_2-\frac{a_4}{a_3}\bigg)$$ $$I=\frac{2u^{3/2}}{3a_3^2}+\frac{2u^{1/2}}{a_3^2}(a_2a_3-a_4)$$ $$I=\frac{2\sqrt{a_3x+a_4}}{3a_3^2}\bigg(a_3(x+3a_2)-2a_4\bigg)+C$$ All that remains is for you to plug in your constants.

Answer 6

Consider the general case of $$I=\int \frac{a x+b}{\sqrt{c x+d}}\,dx$$ Let $$cx+d=t^2 \implies x=\frac{t^2-d}{c}\implies dx= \frac{2 t}{c}\,dt$$ making $$I=\int \left(\frac{2 (b c-a d)}{c^2}+\frac{2 a }{c^2}t^2 \right)\, dt=\frac{2 (b c-a d)}{c^2}t+\frac{2 a }{3c^2}t^3+K$$

Answer 7

Another direct way could be partial integration: $$\begin{eqnarray*} \int\frac{2x+1}{\sqrt{x+3}}dx & = & \int \underbrace{(2x+1)}_{u}\cdot \underbrace{\frac{1}{\sqrt{x+3}}}_{v'}dx \\ & = & (2x+1)\cdot 2\sqrt{x+3} - \int 2 \cdot 2 \sqrt{x+3} \; dx \\ & = & 2 (2x+1)\cdot \sqrt{x+3} - \frac{8}{3}(x+3)\sqrt{x+3} \: \: ( + C )\\ & = & \frac{2}{3}(2x-9)\sqrt{x+3}\: \: ( + C ) \end{eqnarray*}$$

Answer 8

I am quite surprised that nobody yet, suggested, substituting $x+3 = u^2$, this would make your integral very easy to solve.

You will get $dx = 2u.du$, and this $u$ will be cancelled by one in denominator and now you can just substitute $x$ in $2x+1$ with $u^2-3$, and you're pretty much ready to get your answer.