Integration different result different techniques using change of variable

Question

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 $1/x$.

Answer 1

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

$$\frac12\int \frac{du}{u}$$

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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

$$\frac12\int \frac{du}{u}$$

Answer 2

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 ${1\over 2}{ln(\sqrt x+{5\over4})}+c$

Answer 3

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 $$\frac{1}{2}\int\frac{du}{u} = \frac{1}{2}\ln{u}+C = \frac{\ln(4\sqrt{x}+5)}{2}+C$$

Answer 4

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: $$I = \frac12 \ln (4\sqrt x +5) $$

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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.

Answer 5

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.

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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.