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I have problems trying to integrate this. No matter what I try, I still get the question wrong.

$$\int \frac{6}{2x+x\sqrt{x}}dx$$

So this is what I did

$$u = \sqrt{x} \tag{substitution}$$

$$u^2 = x$$

$$2u du=dx$$

Reworking the integral,

$$\int \frac{(6)(2u)}{2u^2+u^2u}du$$

$$=12\int \frac{1}{u(u+2)}du$$

Using partial fraction decomposition

$$\frac{A}{u} + \frac{B}{u+2}=\frac{1}{u(u+2)}$$

$$A(2+u) +B(u) = 1$$

After solving for $A$ and $B$ here

  • When $u=0, A=\frac{1}{2}$

  • When $u=-2, B=\frac{-1}{2}$

Thus my integral becomes this,

$$=12\int\frac{1}{2u}-\frac{1}{2u+4}du$$

And after integration

$$=12\left[\frac{1}{2}\ln|u|-\frac{1}{2}\ln|u+2|\right]+C$$

Substituting back the value of $u$,

$=6\big[\ln|\sqrt{x}|-\ln|\sqrt{x}+2|\big]+C$

This should be my final answer but for some reason it is telling me that I am wrong. Is there any problem along the way?

P/S: I really tried to learn the coding method for the signs, hoping it would make everyone's life easier when anyone tries to help me with this.

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    $\begingroup$ The integral of $\frac{1}{2u}$ is not $\ln(|2u|)$. It is $\frac{1}{2}\ln(|u|)$. (By the way, at the end you do not need absolute value signs.) Thanks for writing out detail, it made analyzing easier. $\endgroup$ – André Nicolas May 12 '16 at 4:32
  • $\begingroup$ $6ln(|\sqrt{x}|)-6ln(|\sqrt{x}+2|)+C$ Is still wrong $\endgroup$ – Jowie Tan May 12 '16 at 4:40
  • $\begingroup$ you have many typo. $$=\int \frac{(6)(2u)}{2u^2+u^2u}du$$ is not correct. It can be $$=\int \frac{(6)(2u)}{2u^2+u}du$$ . The solution will be $$6 \ln (2\sqrt x+1)+C$$ $\endgroup$ – user115350 May 12 '16 at 5:08
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If we make the substitution $u=\sqrt{x}$, we arrive at $$\int \frac{12u\,du}{2u^2+u^3},$$ which simplifies to $$\int \frac{12\,du}{2u+u^2}.$$ Using partial fractions, we find that this is $$\int \frac{6}{u}\,du-\int\frac{6}{u+2}\,du.$$ Integrate. We end up with $$6\ln(|u|)-6\ln(|u+2|)+C.$$ Since $\sqrt{x}$ is non-negative, we can write the final answer as $$6\ln\left(\frac{\sqrt{x}}{\sqrt{x}+2}\right)+C.$$

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  • $\begingroup$ Can you solve it without d? Its confusing $\endgroup$ – Jowie Tan May 12 '16 at 5:07
  • $\begingroup$ Here $d$ is an unspecified constant. Please check carefully to see whether the integral asked for is exactly the one you wrote down. The choice of $d$ as the name of the constant is a little weird, since it invites confusion with the $d$ in $dx$ or $du$. $\endgroup$ – André Nicolas May 12 '16 at 5:10
  • $\begingroup$ Really sorry I made a mistake with the question. There was a missing x that didnt put up on the question. That would make my u substitution a bit more meaningful $\endgroup$ – Jowie Tan May 12 '16 at 5:14
  • $\begingroup$ If you edit your question so that it is exactly what you want, I can probably modify the answer to reflect the change. My current answer calculates the integral exactly as it is written (twice). $\endgroup$ – André Nicolas May 12 '16 at 5:17
  • $\begingroup$ You have changed it in one place. Is the $d$ supposed to be there, or is it a typo? $\endgroup$ – André Nicolas May 12 '16 at 5:19
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$$\int\frac{6}{2x+d\sqrt{x}}\space\text{d}x=6\int\frac{1}{2x+d\sqrt{x}}\space\text{d}x=$$


Substitute $u=\sqrt{x}$ and $\text{d}u=\frac{1}{2\sqrt{x}}\space\text{d}x$:


$$12\int\frac{u}{2u^2+du}\space\text{d}u=12\int\frac{1}{2u+d}\space\text{d}u=$$


Substitute $s=2u+d$ and $\text{d}s=2\space\text{d}u$:


$$6\int\frac{1}{s}\space\text{d}s=6\ln\left|s\right|+\text{C}=6\ln\left|2u+d\right|+\text{C}=6\ln\left|2\sqrt{x}+d\right|+\text{C}$$

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