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Im a newish calc 1 student and i have been trying to solve the integral:

$$\int\dfrac{1}{4x^2 + 9} dx $$

I was able to solve the integral

$\int\frac{1}{x^2 + 1} = \arctan(x) + C$, by doing the method of $x = \tan(u)$ but with the factor of $4$ in the $x^2$ term and +$9$ I'm unsure how to go about solving this, when I put it into wolfram alpha it spits out

$$\frac{1}{6} \arctan\left(\frac{2x}{3}\right) + C$$

But I'm unsure how its came to this answer and cant find any explanations online. I'm not as interested in the actual answer to this specific question as I am how the difference in the terms leads to the different answer

Any help would be greatly appreciated.

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    $\begingroup$ Hint: $4x^2 + 9 = 4\left( x^2 + \frac{9}{4} \right).$ $\endgroup$ Commented Jul 20, 2023 at 16:35

4 Answers 4

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You have an antiderivative for $1/(x^2+1).$ So if you can change your integrand into that form, you'll have the problem solved. So there are two changes to make. You need to change the $9$ into a $1$ and the $4$ into a $1$.

So step one is to factor $1/9$ clear out of the integral:

$$\frac{1}{9} \int \frac{1}{\frac{4}{9}x^2+1}\,dx.$$

Then notice that $\frac{4}{9}x^2 = \left(\frac{2x}{3}\right)^2$, so substitute $u = \frac{2x}{3}.$

These steps will be the same for similar integrals. If the quadratic expression has a linear term then first complete the square, then proceed as above.

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    $\begingroup$ I like this answer because it explains how to get from where the student is to where they know what to do. $\endgroup$ Commented Jul 20, 2023 at 17:21
  • $\begingroup$ The "substitute u = 2x/3" felt like it came out of nowhere in the above answer but this explanation helps a lot, thanks! $\endgroup$
    – Crogmcrob
    Commented Jul 20, 2023 at 17:26
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There is a well known generalized result,

$$\boxed{\int \frac{1}{x^2+a^2}\,dx=\frac{1}{a}arctan\left(\frac{x}{a}\right)+c}$$

For your case, factor out 4 and proceed as follows;

$$I=\frac{1}{4} \int \frac{1}{x^2+\left(\frac{3}{2}\right)^2}\,dx$$

Compare and put, $a=\frac{3}{2}$,

$$I=\frac{1}{6}arctan\left(\frac{2x}{3}\right)+c$$

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    $\begingroup$ Could the downvoter please suggest any required changes? $\endgroup$ Commented Jul 20, 2023 at 17:48
  • $\begingroup$ Imo nothing's wrong here...I did +1 $\endgroup$ Commented Jul 20, 2023 at 18:32
  • $\begingroup$ @MathStackexchangeIsNotSoBad Thanks a lot but it leaves me confused of where the mistake is.... $\endgroup$ Commented Jul 21, 2023 at 4:47
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Substitute $u=\frac{2x}{3}$ to get $dx=\frac32du$ Now rewrite the integral as $$\int\frac{3}{2(9u^2+9)}du$$ I hope you can continue after this.

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First of all take 4 common from the denominator and then substitute x=(3tan(y))/2. Then you will sec²y in the numerator and denominator both and then you have to cancel the term sec²y in both numerator and denominator.The final answer will be 1/6 (arctan(2x/3))+C.

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  • $\begingroup$ For some basic information about writing mathematics at this site see, e.g., here, here, here and here. $\endgroup$ Commented Jul 20, 2023 at 17:14

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