I tried many times to integrate this function using integration by parts, substitution as $x^2+1=t$ without any conclusion... is it integrable? If it is integrable, how can I integrate it? Thank you in advance
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- 3$\begingroup$ It is integrable, but that doesn't mean it has an antiderivative expressable by elementary functions and according to WA, it does not. $\endgroup$Git Gud– Git Gud2014-03-14 14:29:31 +00:00Commented Mar 14, 2014 at 14:29
- $\begingroup$ hypergeometric functions? nope. $\endgroup$Guy– Guy2014-03-14 14:51:43 +00:00Commented Mar 14, 2014 at 14:51
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3 Answers
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1 The excerpt below is from G. H. Hardy's The Integration of Functions of a Single Variable [see p. 51 in the Project Gutenberg version].
One particular class of integrals which is of especial interest is that of the binomial integrals
$$\int x^m \left(ax^n + b\right)^{p} \, dx,$$
where $m,$ $n,$ $p$ are rational. Putting $ax^n = bt,$ and neglecting a constant factor, we obtain an integral of the form
$$\int t^q (1 + t)^{p} \, dt,$$
where $p$ and $q$ are rational. If $p$ is an integer, and $q$ a fraction $r/s,$ this integral can be evaluated at once by putting $t = u^{s},$ a substitution which rationalises the integrand. If $q$ is an integer, and $p = r/s,$ we put $1 + t = u^{s}.$ If $p+q$ is an integer, and $p = r/s,$ we put $1 + t = tu^{s}.$
It follows from Tschebyschef's researches (to which references are given in Appendix I) that these three cases are the only ones in which the integral can be evaluated in finite form.
For the integral above,
$$\int \left(x^2 + 1\right)^{\frac{1}{3}} dx,$$
we have $m=0,$ $a=1,$ $n=2,$ $b=1,$ and $p = \frac{1}{3}.$ Thus, we use the substitution $x^2 = t.$ For this substitution we have $dx = \frac{1}{2} x^{-1} dt = \frac{1}{2} t^{-\frac{1}{2}}dt,$ which gives
$$\frac{1}{2} \int t^{-\frac{1}{2}}(1 + t)^{\frac{1}{3}} dt$$
Therefore, for the integral above, we have $p = \frac{1}{3}$ and $q = -\frac{1}{2}.$ These values of $p$ and $q$ do not belong to one of the three solvable cases that Hardy singled out, and thus it follows that the integral cannot be evaluated "in finite form".
- $\begingroup$ Maybe we should have a badge for being the "selected best answer" in conjunction with a negative point total? For some reason I went from 0 votes to one down vote in the past day, and additionally I have no idea what is so lacking in my answer. $\endgroup$Dave L. Renfro– Dave L. Renfro2014-03-19 16:20:39 +00:00Commented Mar 19, 2014 at 16:20
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Mathematica suggests $$ \int dx(x^2+1)^{1/3}=\frac{3}{5}x(x^2+1)^{1/3}+\frac{2x}{5} \cdot \ _2F_1[\frac{1}{2},\frac{2}{3},\frac{3}{2},-x^2] $$ where the Hypergeometric function is given by $$ _2F_1(a,b,c;x)=\sum_{k=0}^\infty \frac{(a)_k (b)_k (c)_k x^k}{(c)_k k!}. $$ To answer your questions, it is integrable, however the result is in terms of special functions (HYpergeomtric functions) and NOT elementary functions such as trigonometric, logarithmic, exponential, etc. Methods like integration by parts (and other methods) still work whether you're dealing with special functions or elementary functions however. If you are trying to learn integration, perhaps it is more wise at first to try something integrable in terms of elementary functions such as $$ \int dx(x^2+1)^{1/2}. $$
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Let $u=(x^2+1)^\frac{1}{3}$ ,
Then $x=(u^3-1)^\frac{1}{2}$
$dx=\dfrac{3u^2}{2(u^3-1)^\frac{1}{2}}du$
$\therefore\int(x^2+1)^\frac{1}{3}~dx=\int\dfrac{3u^3}{2(u^3-1)^\frac{1}{2}}du$
Case $1$: $|u|\leq1$ , i.e. $\left|(x^2+1)^\frac{1}{3}\right|\leq1$
Then $\int\dfrac{3u^3}{2(u^3-1)^\frac{1}{2}}du$
$=\int\dfrac{3u^3}{2i(1-u^3)^\frac{1}{2}}du$
$=-\int\dfrac{3u^3i}{2}\sum\limits_{n=0}^\infty\dfrac{(2n)!u^{3n}}{4^n(n!)^2}du$
$=-\int\sum\limits_{n=0}^\infty\dfrac{3i(2n)!u^{3n+3}}{2^{2n+1}(n!)^2}du$
$=-\sum\limits_{n=0}^\infty\dfrac{3i(2n)!u^{3n+4}}{2^{2n+1}(n!)^2(3n+4)}+C$
$=-\sum\limits_{n=0}^\infty\dfrac{3i(2n)!(x^2+1)^{n+\frac{4}{3}}}{2^{2n+1}(n!)^2(3n+4)}+C$
Case $2$: $|u|\geq1$ , i.e. $\left|(x^2+1)^\frac{1}{3}\right|\geq1$
Then $\int\dfrac{3u^3}{2(u^3-1)^\frac{1}{2}}du$
$=\int\dfrac{3u^3}{2u^\frac{3}{2}(1-u^{-3})^\frac{1}{2}}du$
$=\int\dfrac{3u^\frac{3}{2}}{2}\sum\limits_{n=0}^\infty\dfrac{(2n)!u^{-3n}}{4^n(n!)^2}du$
$=\int\sum\limits_{n=0}^\infty\dfrac{3(2n)!u^{\frac{3}{2}-3n}}{2^{2n+1}(n!)^2}du$
$=\sum\limits_{n=0}^\infty\dfrac{3(2n)!u^{\frac{5}{2}-3n}}{2^{2n+1}(n!)^2\left(\dfrac{5}{2}-3n\right)}+C$
$=-\sum\limits_{n=0}^\infty\dfrac{3(2n)!}{4^n(n!)^2(6n-5)u^{3n-\frac{5}{2}}}+C$
$=-\sum\limits_{n=0}^\infty\dfrac{3(2n)!}{4^n(n!)^2(6n-5)(x^2+1)^{n-\frac{5}{6}}}+C$