# 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 $1/x$.

• math.meta.stackexchange.com/questions/5020/… – 5xum Jan 3 '18 at 10:35
• I translated your expressions to LaTeX, please check I didn't change what you meant. – Jean-Claude Arbaut Jan 3 '18 at 11:58
• @Jean-ClaudeArbaut in the dx it's 2/sqrt(x) not 2*sqrt(x) – john Jan 3 '18 at 12:00
• Is it better now? – Jean-Claude Arbaut Jan 3 '18 at 12:04
• @Jean-ClaudeArbaut yes – john Jan 3 '18 at 12:05

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

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

• can you check my question again... I edited it... and tell me where am I mistaking – john Jan 3 '18 at 11:54
• @john I added another answer. I feel you didn't do the substitution correctly, so I urge you to write everything as carefully as you can. – Jean-Claude Arbaut Jan 3 '18 at 12:03

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

• yes but why my result is different? – john Jan 3 '18 at 10:40
• Do you mean that you replace x with 2/sqrt(x)? – Mostafa Ayaz Jan 3 '18 at 10:43
• yes inside the dx – john Jan 3 '18 at 10:44
• I guess the '5' constant in the denominator bans you from forwarding in proof outline.... – Mostafa Ayaz Jan 3 '18 at 10:47
• I still don't get where my mistake is – john Jan 3 '18 at 10:53

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

• I put the 2/√x under the dx and it becomes d(2√x) because the first derivate of 4√x+5 is 2/√x – john Jan 3 '18 at 10:52
• Then I think you must put 4sqrt(x)+5 under d(.) and then your calculations gonna be true... – Mostafa Ayaz Jan 3 '18 at 11:16

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.

• well not exactly... I got 1/x ... First I substituted 4√x+5 =u and I get S 1/√x d(2/√x)... then i multiplied my integral by 2/2 so i get 1/2 S 2/√x d(2/√x) and I used the basic power rule – john Jan 3 '18 at 11:30
• @john Then your mistake is easy to spot: When you substitute $4\sqrt{x}+5=u$, you are left with $1/u$, not $1/\sqrt{u}$. – Jean-Claude Arbaut Jan 3 '18 at 11:33
• @Jean-ClaudeArbaut can't I just multiply my 1/sqrt(x) by 2/2 ?? and it gets 2/sqrt(x) which is same as the u? – john Jan 3 '18 at 11:35
• No, if you multiply $1/\sqrt{x}$ by $2/2$, it does not become $2/\sqrt{x}$. And no, if $u=4\sqrt{x}+5$, then $2/\sqrt{x}$ is not equal to $u$. However, you can write $du=2dx/\sqrt{x}$. And since there is already a $1/\sqrt{x}$ in your integrand, that simplifies cleanly. – Jean-Claude Arbaut Jan 3 '18 at 11:38
• @Jean-ClaudeArbaut I don't get it... in class we studied that I can multiply it and it becomes 1/2 S 2/√x du ... the first derivative of u is 2/√x... so my u =2/√x – john Jan 3 '18 at 11:41

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.

• anyway as my question... Where is my mistake – john Jan 3 '18 at 12:11
• @john What is your substitution? You didn't even write one. I'm trying to help, but you have to be accurate. If you substitute $x=g(u)$, what is your $g$? – Jean-Claude Arbaut Jan 3 '18 at 12:23