Calculate integral $\int \frac{x^2+1}{\sqrt{x^3+3}}dx$ This was my exam question. I've tried many online math solvers and math programs but none were able to solve. If anybody has an answer would be helpful. Thanks
Asked
Active
Viewed 602 times
6
-
6I'll bet a dollar that the $x^3+3$ was supposed to be $x^3+3x.$ – B. Goddard Jan 17 '18 at 11:36
-
Not sure if it is a typo: were you asked to find the antiderivative? If so, I would be surprised if an answer in closed form exists! – Szeto Jan 17 '18 at 11:37
-
yes it was antiderivative, if no answer i believe it's a typo – alynurly Jan 17 '18 at 11:39
-
I got $\frac{x\sqrt{x^2+3}-\ln {(|2x+2\sqrt{x^2+3}|)}}{2} + c$. – Landuros Jan 17 '18 at 11:40
-
2@Landuros In the question, it's $\sqrt{x^\color{red}3+3}$. – Jan 17 '18 at 11:50
-
2probably a typo, it leads to an elliptic function – Dr. Sonnhard Graubner Jan 17 '18 at 11:56
-
It must be a typo, because the result of http://www.wolframalpha.com/input/?i=integrate+(x%5E2+%2B+1)+%2F+sqrt(x%5E3+%2B+3) is for sure not a part of an exam. – user90369 Jan 17 '18 at 14:40
-
@ProfessorVector Whoopsies :). – Landuros Jan 18 '18 at 01:45
-
@Landuros But you are probably right, or it's B. Goddard. There is certainly a typo, but the OP has to check this with his teacher and tell us. – Jean-Claude Arbaut Jan 18 '18 at 16:52
-
Function $y=\frac{x^2+1}{\sqrt{x^3+3}}$ has a minimum at $x=0, y=1/\sqrt 3$. $x→∞ ⇒ y→∞ $ and $ if x→-∞ ⇒ y→-∞$. The solution is the integral of a series resulted from$(x^2+1)(binomial expansion of (x^3+3)^{-1/2}$. – sirous Jan 19 '18 at 15:28
2 Answers
5
$$I = \int\frac{x^2+1}{\sqrt{x^3+3}}dx=\underbrace{\int \frac{x^2}{\sqrt{x^3+3}}dx}_{I_1}+\underbrace{\int \frac{1}{\sqrt{x^3+3}}dx}_{I_2}$$
$$I_1 = \int \frac{x^2}{\sqrt{x^3+3}}dx$$
u-substitution $u=x^3$
$$\int \frac{1}{3\sqrt{u+3}}du =\frac{1}{3}\int \frac{1}{\sqrt{u+3}}du$$
u-substitution $v=u+3$
$$I_1 = \frac{1}{3} \int \frac{1}{\sqrt{v}}dv = \frac{1}{3}\int v^{-\frac{1}{2}}dv = \frac{1}{3}\frac{v^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C$$
Revert the substitutions $v=u+3,u=x^3$
$$I_1 = \frac{2}{3}\sqrt{x^3+3} + C$$
Now
\begin{align}
I_2 &= \int \frac{1}{\sqrt{x^3+3}}dx\\
&= \int \frac{1}{\sqrt{3\left(\frac{x^3}{3}+1\right)}}dx\\
&= \int \frac{1}{\sqrt{3\left(\left(\frac{x}{\sqrt[3]{3}}\right)^3+1\right)}}dx
\end{align}
Set $u = \frac{x}{\sqrt[3]{3}} \implies dx = \sqrt[3]{3}\, du$, then
\begin{align}
I_2&= \frac{\sqrt[3]{3}}{\sqrt{3}}\int \frac{1}{\sqrt{u^3+1}}du\\
\end{align}
See : $\int\frac{1}{\sqrt{x^3+1}}$
\begin{align}
I_2&= \frac{\sqrt[3]{3}}{\sqrt{3}}\left[\frac1{\sqrt[4]{3}}F\left(\arccos\left(\frac{2\sqrt{3}}{1+\sqrt{3}+u}-1\right)\mid\frac{2+\sqrt{3}}{4}\right)\right]+C\\
\end{align}
Revert the original substitution $u = \frac{x}{\sqrt[3]{3}} $:
\begin{align}
&= \frac{\sqrt[3]{3}}{\sqrt{3}\sqrt[4]{3}}F\left(\arccos\left(\frac{2\sqrt{3}}{1+\sqrt{3}+\frac{x}{\sqrt[3]{3}}}-1\right)\mid\frac{2+\sqrt{3}}{4}\right)+C\\
\end{align}
And
$$I = \frac{2}{3}\sqrt{x^3+3} + \frac{\sqrt[3]{3}}{\sqrt{3}\sqrt[4]{3}}F\left(\arccos\left(\frac{2\sqrt{3}}{1+\sqrt{3}+\frac{x}{\sqrt[3]{3}}}-1\right)\mid\frac{2+\sqrt{3}}{4}\right)+C$$
Darío A. Gutiérrez
- 3,863
1
I have found a solution in terms of a hypergeometric function for positive $x$: $$ I = \frac{2}{3\sqrt{x}}\left(\sqrt{1+\frac{3}{x^3}}(1+x^2)-4\;_2F_1\left(-\frac{1}{2},\frac{1}{6},\frac{7}{6},-\frac{3}{x^3}\right)\right) +C $$ and verified the derivative equals the integrand (for positive $x$), and in fact for general $a$, $$ \int \frac{x^2+1}{\sqrt{x^3+a}}dx $$ is given by $$ I = \frac{2}{3\sqrt{x}}\left(\sqrt{1+\frac{a}{x^3}}(1+x^2)-4\;_2F_1\left(-\frac{1}{2},\frac{1}{6},\frac{7}{6},-\frac{a}{x^3}\right)\right) + C $$
Benedict W. J. Irwin
- 4,321