I have tried this:

$$\frac1{x^2+x+1} = \frac1{\left( (x+\frac12)^2+\frac34\right)}$$

Now $u = x+\frac12$

$$\frac1{ u^2+\frac34 }$$ Now multiply by $ \frac34$

$$\frac1{ \frac43 u^2 + 1}$$

Now put the $\frac43$ outside the integral

$$\frac34 \int \frac1{u^2+1}\,du=\frac34\arctan(u)=\frac34\arctan(x+1/2)$$

But the result is not the same result calculated by computers.

What did I do wrong?

Could someone please help me with this?

I don't know where my wrong calculation is. The way should be correct to get to the result.


So now

$\int \frac{1}{x^2+x+1}=\int \frac{1}{(x+\frac{1}{2})^2}= \int \frac{4}{3} \frac{1}{\frac{4}{3}(x+\frac{1}{2})^2+1}$ u=x^2+1/2 $\int \frac{4}{3} \frac{1}{\frac{4}{3}(u)^2+1}$

$\int \frac{4}{3} \frac{1}{\frac{4u^2}{3}+1}$

$\int \frac{4}{3} \frac{1}{\frac{2u^2}{\sqrt{3}}+1}$

Now it is:

$\frac{4}{3} \int \frac{1}{\frac{2u^2}{\sqrt{3}}+1}$

$=\frac{4}{3} * arctan(2*(x^2+1/2)/(\sqrt{3}))$

Why is this still not the same as the computer calculated solution?

  • 2
    $\begingroup$ Please use MathJax to render your math equations $\endgroup$ – Shubhrajit Bhattacharya Jul 30 '20 at 23:06
  • 3
    $\begingroup$ $\dfrac1{u^2+\frac34}$ is not the same as $\dfrac1{\frac43u^2+1}$, which is not the same as $\dfrac34\dfrac1{u^2+1}$ (try $u=0$, for example) $\endgroup$ – J. W. Tanner Jul 30 '20 at 23:27
  • $\begingroup$ @ShubhrajitBhattacharya how do I use mathjax here? I just know LaTex $\endgroup$ – Rapiz Jul 30 '20 at 23:58
  • $\begingroup$ @Rapiz go here math.meta.stackexchange.com/questions/5020/… $\endgroup$ – Shubhrajit Bhattacharya Jul 31 '20 at 0:08
  • $\begingroup$ Also MathJax is almost same as LaTeX except some differences that you will come across if you continue to use MathJax $\endgroup$ – Shubhrajit Bhattacharya Jul 31 '20 at 0:10

Note that $$\frac{1}{u^2+3/4}=\frac{1}{\frac34((2u/\sqrt 3)^2+1)}=\frac43 \frac{1}{v^2+1}$$

where $v=2u/\sqrt 3$. Can you finish now?

  • 2
    $\begingroup$ Again we have $$\frac1{u^2+3/4}=\frac{4/3}{4/3}\frac1{u^2+3/4}=\frac{4/3}{4u^2/3+1}=\frac43 \frac1{(2u/\sqrt 3)^2+1}$$ $\endgroup$ – Mark Viola Jul 30 '20 at 23:17
  • 3
    $\begingroup$ @Rapiz Multiplying top and bottom by $4/3$ gives $\frac{4}{3} \frac{1}{(4/3) u^2+1}$. You can't just multiply the bottom alone by $4/3$, that would change the result. $\endgroup$ – Ian Jul 30 '20 at 23:18
  • 2
    $\begingroup$ @Rapiz No, because now you've changed the integrand, that's what I was trying to say. If you want to multiply the bottom by $4/3$ (so as to make the $3/4$ into a $1$ without factoring the way that Mark Viola did) then you have to do the same to the top. $\endgroup$ – Ian Jul 30 '20 at 23:22
  • 2
    $\begingroup$ Is the ratio $\frac{a}{b}=\frac{a}{(4/3)b}$? It isn't. But we can write $$\frac ab=\frac{(4/3)a}{(4/3)b}$$ $\endgroup$ – Mark Viola Jul 30 '20 at 23:23
  • 2
    $\begingroup$ You cannot simply multiply the denominator by $4/3$ and not do the same to the numerator. The integrand IS a ratio. It is $\frac1{u^2+3/4}$. So, here $a=1$ and $b=u^2+3/4$. Now how can we preserve equality? We can multiply by $1=\frac{4/3}{4/3}$. $\endgroup$ – Mark Viola Jul 30 '20 at 23:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.