# a way to integrate: $\int (\sqrt{x} +3)/(2+ x^ \frac{1}{3}) dx$

Im looking for a way to integrate: $$\int \frac{ \sqrt{x} +3}{2+ x^ \frac{1}{3}} dx$$ that would make it efficient and not too difficult...

Any suggestions?

• I haven't tried, but what about changing x = y^6? Does it get better? Jan 7, 2015 at 21:36
• its not a bas suggestion, problem is you still need to replace $dx$ properly which might be a bit problematic in that course. Jan 7, 2015 at 21:56
• Well, if $x=y^6$ then $dx=6y^5\,dy$. Jan 7, 2015 at 21:58
• @BCLC, you would actually wind up integrating ${(u-2)^{7/2}\over u}du$.... Jan 8, 2015 at 0:02
• @BCLC, $x=(u-2)^3$ means $dx=3(u-2)^2du$. Jan 8, 2015 at 1:16

The substitution $x = t^6$ works. The integral becomes

$$\int \frac{t^3 + 3}{t^2 + 2} \,6t^5\,dt = 6 \int \frac{t^8 + 3t^5}{t^2 + 2} dt = 6\int \left( t^6 - 2t^4 + 3t^3 + 4t^2 - 6t - 8 + \frac{12t+16}{t^2+2} \right) \, dt$$

You can take it from here

• how did you get from position two to three? Jan 9, 2015 at 19:58
• I used long division. Jan 12, 2015 at 20:43

Let $x=u^6$, $dx=6u^5du$. \begin{align*} &\int\frac{\sqrt x+3}{2+\sqrt[3]x}\,dx\\ =&\int\frac{u^3+3}{u^2+2}6u^5\,du\\ =&6\int\frac{u^8+3u^5}{u^2+2}\,du\\ =&6\int\left(u^6-2u^4+3u^3+4u^2-6u-8+\frac{12u+16}{u^2+2}\right)\,du\\ =&6\int\left(u^6-2u^4+3u^3+4u^2-6u-8+6\frac{2u}{u^2+2}+8\frac{1}{\left(\frac{\sqrt2}{2}u\right)^2+1}\right)\,du\\ =&\frac67u^7-\frac{12}{5}u^5+\frac92u^4+8u^3-18u^2-48u+36\ln\left(u^2+2\right)+48\sqrt2\arctan\left(\frac{\sqrt2}{2}u\right)+C\\ =&\frac67x^{7/6}-\frac{12}{5}x^{5/6}+\frac92x^{2/3}+8x^{1/2}-18x^{1/3}-48x^{1/6}+\\ &36\ln\left(x^{1/3}+2\right)+48\sqrt2\arctan\left(\frac{\sqrt2}{2}x^{1/6}\right)+C \end{align*}

• Shouldn't it rather be $u^8+3u^5$ in place of $u^8+3$ at line 3 of the equations? Jan 7, 2015 at 23:19
• Yes, thanks for pointing it out :)
– slo
Jan 7, 2015 at 23:23