Integration different result different techniques using change of variable I am currently integrating $$\int \frac{1}{\sqrt{x}\cdot(4\sqrt{x}+5)}dx$$
When I change the variable in the $dx$ to $\frac{2}{\sqrt{x}}$ I can multiply the integral by $\frac{2}{2}$ (which is $1$) (multiply only the numerator by $2$ and leave the $\frac{1}{2}$ outside the integral) that way I can use the power rule for integrals and after simplification I am left with $\frac{1}{x}$... however the online calculators give different result:
$$\frac{\ln(4\sqrt{x}+5)}{2}$$
What I did is
$$\int \frac{1}{\sqrt{x}}  d(\frac{2}{\sqrt x})$$
Then I multiplied by $\frac22$ and I get
$$\frac12 \int \frac{2}{\sqrt x}d(\frac{2}{\sqrt x})$$
And I use the power rule...
After simplification I get $1/x$.
 A: Why don't you try $x=t^{2}$? Then the integral simplifies to $\int{2\over 4t+5}dt$ and is equal to ${1\over 2}{ln(t+{5\over4})}+c$ or ${1\over 2}{ln(\sqrt x+{5\over4})}+c$
A: I didn't understand what you mean by changing in the variable $dx$ to $\frac{2}{\sqrt{x}}$ but I think what you should have done is $u = 4\sqrt{x}+5$ then $du = \frac{2}{\sqrt{x}}dx$. Then the expression becomes 
$$\frac{1}{2}\int\frac{du}{u} = \frac{1}{2}\ln{u}+C = \frac{\ln(4\sqrt{x}+5)}{2}+C$$
A: Let's do it step by step.
$$\int \frac{1}{\sqrt{x}\cdot(4\sqrt{x}+5)}dx=\int \frac{1}{4\sqrt{x}+5}\frac{dx}{\sqrt{x}}$$
Let's do the substitution $4\sqrt{x}+5=u$. Then $du=\dfrac{2dx}{\sqrt{x}}$, and the $\dfrac{dx}{\sqrt{x}}$ already here in your integral becomes $\dfrac{du}{2}$, while $4\sqrt{x}+5$ becomes $u$.
The integral thus becomes
$$\frac12\int \frac{du}{u}$$

If the simplification looks too "fast", you can do it this way:
Since $du=\dfrac{2dx}{\sqrt{x}}$, you have $dx=\frac12\sqrt{x}du$, then the integral becomes
$$\int \frac1{\sqrt{x}}\cdot\frac{1}{4\sqrt{x}+5}\cdot \frac12\cdot\sqrt{x}du$$
$$\int \frac1{\sqrt{x}}\cdot\frac{1}u\cdot \frac12\cdot\sqrt{x}du$$
And the two $\sqrt{x}$ simplify, to leave
$$\frac12\int \frac{du}{u}$$
A: Basically, this is what you must have done first: $$I = \frac12 \int \frac {2}{\sqrt x} \times \frac {1}{4\sqrt x +5}\, dx $$ 
Now, notice as the first derivative of $4\sqrt x +5 $ is $\frac {2}{\sqrt x}$, we will get: $$I = \frac12 \int \frac {\mathrm d (4\sqrt x +5)}{4\sqrt x +5} $$ This is a very common form of an integral: $$ \int \frac {1}{x}\, dx = \log x + c \tag 1 $$ Using this result, we will get: $$I = \frac12 \ln (4\sqrt x +5) $$

What you must have done is: you could have substituted $4\sqrt x +5 = t$ to get the form $(1) $, but you must have just left it like that, without integrating. 
Otherwise, you will get a result of $\frac1 {x} $ only on integrating $-\frac {1}{x^2} $, which has no place in our integral here.
A: When you do integration by substitution in
$$\int f(x)dx$$
You replace the integration variable $x$ with some function of another variable, say $u$ (for instance $x=g(u)$), with some conditions on the function $g$. Then the integral becomes
$$\int f(g(u)) g'(u)du$$
Please write down which function $g$ you used.
You seem to be mixing several concepts, and this makes very unclear what you meant. Get back to the definition.

The informal way to do substitution is to write $x=g(u)$ hence $dx=g'(u)du$, or if you prefer $\dfrac{dx}{du}=g'(u)$. Then, in the original integral, you replace $dx$ with $g'(u)du$, and any other instance of $x$ with $g(u)$.
It gets quickly to a result, but I don't like very much working with $d$, without a proper definition. However, it's exactly equivalent to the more formal way, if you do it correctly.
