Krdytkyu

I'm trying to solve this:

$$intfrac{2x+1}{sqrt{x+3}}dx$$

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

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.

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

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

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

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

What's next?

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.

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 $x=frac{u-a_4}{a_3}$
$$I=frac1{a_3}int frac{frac{a_1}{a_3}u-frac{a_4}{a_3}+a_2}{u^{1/2}}du$$
$$I=frac1{a_3^2}intfrac{u}{u^{1/2}}du+frac1{a_3}bigg(a_2-frac{a_4}{a_3}bigg)intfrac{du}{u^{1/2}}$$
$$I=frac{2u^{3/2}}{3a_3^2}+frac{2u^{1/2}}{a_3}bigg(a_2-frac{a_4}{a_3}bigg)$$
$$I=frac{2u^{3/2}}{3a_3^2}+frac{2u^{1/2}}{a_3^2}(a_2a_3-a_4)$$
$$I=frac{2sqrt{a_3x+a_4}}{3a_3^2}bigg(a_3(x+3a_2)-2a_4bigg)+C$$
All that remains is for you to plug in your constants.

$$intfrac{2x+1}{sqrt{x+3}}dx=intfrac{2(x+3)-5}{sqrt{x+3}}d(x+3)=intfrac{2u-5}{sqrt u}du$$
$du=dx$ 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
$$intfrac{2x+1}{sqrt{x+3}}dx=intfrac{2u-5}{sqrt u}frac{dx}{du}du=intfrac{2u-5}{sqrt u}du$$

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

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
$$I=int left(frac{2 (b c-a d)}{c^2}+frac{2 a }{c^2}t^2 right), dt=frac{2 (b c-a d)}{c^2}t+frac{2 a }{3c^2}t^3+K$$

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