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I need to learn how to find the definite integral of the square root of a polynomial such as: $$\sqrt{36x + 1}$$ or $$\sqrt{2x^2 + 3x + 7} $$

EDIT: It's not guaranteed to be of the same form. It could be any polynomial that can't be easily factored into squares.

This isn't homework, I'm studying for a final. And for context, I'm finding the arc length of a function.

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    $\begingroup$ Try $u=36x+1$.. $\endgroup$ – vadim123 May 13 '13 at 4:35
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    $\begingroup$ So you want to integrate $\sqrt{P(x)}$. If $P(x)$ is a linear or quadratic polynomial, we can integrate in terms of elementary functions. In general, if the degree of $P(x)$ is $\ge 3$, $\sqrt{P(x)}$ does not have an elementary antiderivative except in quite special cases. $\endgroup$ – André Nicolas May 13 '13 at 4:46
  • $\begingroup$ This is a longshot, but, is there an efficient way to compute it? $\endgroup$ – Scott Jan 3 '18 at 1:50
  • $\begingroup$ @AndréNicolas where can I find the proof of the conclusion? $\endgroup$ – athos Jun 6 '18 at 3:53
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Recall that for $\alpha \neq -1$, we have $$\int (ax+b)^{\alpha}dx = \dfrac1a \cdot \dfrac{(ax+b)^{\alpha+1}}{\alpha+1} + \text{ constant}$$ A way to see the above is as follows. Let $y = ax+b$. We then have $dy = adx$. Hence for $\alpha \neq -1$, $$\int (ax+b)^{\alpha}dx = \int y^{\alpha} \dfrac{dy}a = \dfrac1a \dfrac{y^{\alpha+1}}{\alpha+1} + \text{ constant} = \dfrac1a \dfrac{(ax+b)^{\alpha+1}}{\alpha+1} + \text{ constant}$$ If $\alpha = -1$, we then have $$\int \dfrac{dx}{ax+b} = \dfrac{\log(ax+b)}a + \text{ constant}$$


In general, there is no easy way to get $$\int \sqrt{P(x)} dx,$$ if degree of $P(x)$ is greater than $2$.

If $P(x)$ is linear, i.e., has degree $1$, I have mentioned above how to proceed.

Below we will see how to proceed if $P(x)$ is quadratic, i.e., $$P(x) = ax^2 + bx + c = a ((x+b_1)^2 + c_1).$$ $$b_1=\frac{a}{2b}$$ $$c_1=\frac{c}{a}-b_1^2$$

We now have $$\sqrt{P(x)} = \sqrt{a} \sqrt{(x+b_1)^2 \pm c_1^2}$$ This gives us that $$\int \sqrt{P(x)} dx = \sqrt{a} \int \sqrt{(x+b_1)^2 + c_1} dx$$ and $$\int \sqrt{(x+b_1)^2 + c_1} dx = \dfrac{(b_1+x)\sqrt{P(x)} + c_1 \log \left(b_1 + x + \sqrt{P(x)}\right)}2 + \text{constant}$$

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  • $\begingroup$ @CameronBuie Thanks. Updated. $\endgroup$ – user17762 May 13 '13 at 4:40
  • $\begingroup$ Sorry for not being more explicit, but I need a more general solution. It's not guaranteed to be ax+b $\endgroup$ – notbad.jpeg May 13 '13 at 4:40
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    $\begingroup$ @notbad.jpeg Your question wants the integral of $\sqrt{36x+1}$. This means $\alpha = \dfrac12$, $a = 36$ and $b=1$. $\endgroup$ – user17762 May 13 '13 at 4:41
  • $\begingroup$ I edited the question. $\endgroup$ – notbad.jpeg May 13 '13 at 4:43
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    $\begingroup$ In "we now have", it seems that the c term gains an unnecessary square as in $c_{1}^{2}$ $\endgroup$ – Charlie Oct 4 '17 at 19:59
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For a linear polynomial under the radical a $u$-substitution will do. If you wish to solve this problem for higher order polynomials in the general case then you will be appealing to a lot of algebra, hyperbolic trig functions, and elliptic integrals (which are similar in that they are arclength problems). In short there is no easy short cut to solve this problem for arbitrary degree polynomials.

For degree one polynomials, do as the other answers have advised. For degree two polynomials, like $\sqrt{2x^{2}+3x+7}$, you should employ trigonometric substitutions.

Step 1: Given $\int\sqrt{ax^{2}+bx+c}$ first complete the square into something of the form $k\int\sqrt{\pm u^{2}\pm l}$.

Step 2: Using one of the three substitutions from this article on the matter to get the integral into the form of $k\int\sqrt{f(x)^{2}}$ for some trig function $f(x)$.

Step 3: Remove the radical and solve using known integrals.

If you wish to venture into solving cubics (or higher degree polys) in the general case (i.e. not reducible by a substitution) then do read up on elliptic integrals.

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  • $\begingroup$ user17762's method with logarithms is more concise and general, but the trig function one is often easier to recall on the fly in exams. $\endgroup$ – Kaya May 13 '13 at 5:07
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Using the $u$-substitution $u = 36x + 1$, we have $du = 36dx$. Then, it follows that:

$$\int_{a}^{b} \sqrt{36x+1}dx = \frac{1}{36}\int_{36a+1}^{36b+1}u^{\frac{1}{2}}du = \frac{1}{36}*\frac{2}{3}u^{\frac{3}{2}}|^{36b+1}_{36a+1} = \frac{1}{54}(36x+1)^{\frac{3}{2}}|^{b}_{a}$$

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  • $\begingroup$ This works, but I thought that I could complete the square or something like that. $\endgroup$ – notbad.jpeg May 13 '13 at 4:44
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    $\begingroup$ @notbad.jpeg, the polynomial under the radical is not a quadratic, so completing the square doesn't quite work here. For your more general question, see André Nicolas's comment. $\endgroup$ – Alex Wertheim May 13 '13 at 4:48
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$$\int {\sqrt{36x+1}}\,dx$$

use this formula

$$\int {\sqrt{x}}\,dx=\frac {x^{\frac{3}{2}}}{\frac32}\implies \frac {2x^\frac32}{3}+C$$

so $$\int {\sqrt{36x+1}}\,dx=\frac {2(36x+1)^\frac32}{(3)(36)}$$ $$\frac {(36x+1)^\frac32}{54}+C$$

Q2

$$\int\sqrt{2x^2 + 3x + 7}\,dx $$ $$\int\sqrt{2(x^2 + \frac{3x}{2} + \frac72)}\,dx $$ $$\int\sqrt{2(x^2 + \frac{3x}{2} + \frac72)}\,dx $$ $$\sqrt2\int\sqrt{(x^2 + \frac{3x}{2} + \frac72)}\,dx $$ $$\sqrt2\int\sqrt{(x^2 + \frac{3x}{2} + \frac72+\frac{3^2}{4^2}-\frac{3^2}{4^2})}\,dx $$ $$\sqrt2\int\sqrt{(x^2 + \frac{3x}{2} +\frac{3^2}{4^2}+ \frac72-\frac9{16})}\,dx $$ $$\sqrt2\int\sqrt{(x+ \frac{3}{4})^2 + \frac72-\frac9{16})}\,dx $$ $$\sqrt2\int\sqrt{(x+ \frac{3}{4})^2 + (\frac{\sqrt47}{4})^2}\,dx $$

now use this formula $$\int{\sqrt{x^2+a^2}}\,dx=\frac{x\sqrt{x^2+a^2}}{2}+\frac{a^2\log|x+\sqrt{x^2+a^2}|}{2}+C$$ I hope you can take it now on your own

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  • $\begingroup$ I think that's the same as u-substitution $\endgroup$ – notbad.jpeg May 13 '13 at 5:00
  • $\begingroup$ yes it is same as u substitution.But I do not use subs. if question is in standard form of any basic formula $\endgroup$ – iostream007 May 13 '13 at 5:17

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