Integrate $ \int \sqrt{\frac{x^3-3}{x^{11}}}\ dx$

Question

$$ \int \sqrt{\frac{x^3-3}{x^{11}}}\ dx $$

It seems like substitution could not do it. Is there another way?

Answer 1

Notice, $$\int\sqrt{\frac{x^3-3}{x^{11}}}dx$$

$$=\int\frac{\sqrt{x^3-3}}{x^{11/2}}dx$$

Let, $x^3=3\sec^2\theta\implies 3x^2dx=6\sec^2\theta\tan \theta d\theta$ $$dx=\frac{2\sec^2\theta\tan \theta d\theta}{(3\sec^2\theta)^{2/3}}$$ $$\int\frac{\sqrt{3\sec^2\theta-3}\frac{2\sec^2\theta\tan \theta d\theta}{(3\sec^2\theta)^{2/3}}}{(3\sec^2\theta)^{11/6}}$$ $$=\int\frac{2\sqrt 3\sec^2\theta\tan^2 \theta d\theta}{(3\sec^2\theta)^{11/6}(3\sec^2\theta)^{2/3}}$$ $$=\int\frac{2\sqrt 3\sec^2\theta\tan^2 \theta d\theta}{(3)^{5/2}\sec^5\theta}$$ $$=\frac{2}{9}\int\frac{\tan^2 \theta d\theta}{\sec^3\theta}$$ $$=\frac{2}{9}\int\cos^3\theta \frac{\sin^2 \theta d\theta}{\cos^2\theta}$$

$$=\frac{2}{9}\int\sin^2 \theta\cos\theta d\theta$$ Let $\sin \theta=t\implies \cos \theta d\theta=dt$ $$\frac{2}{9}\int t^2dt=\frac{2}{9}\frac{t^3}{3}+C$$ $$=\frac{2t^3}{27}+C$$ $$=\frac{2\sin^3 \theta}{27}+C$$

Now, substitute $$\sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{1-\frac{1}{\sec^2\theta}}=\sqrt{1-\frac{3}{x^3}}=\sqrt{\frac{x^3-3}{x^3}}$$ $$\iff\sin^3\theta=\left(\frac{x^3-3}{x^3}\right)^{3/2}$$ $$=\frac{2}{27}\left(\frac{x^3-3}{x^3}\right)^{3/2}+C$$

Hence, we get

$$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{\int\sqrt{\frac{x^3-3}{x^{11}}}dx=\frac{2}{27}\left(\frac{x^3-3}{x^3}\right)^{3/2}+C}}$$

Answer 2

$x^3-3=y^2 \Rightarrow x=(y^2+3)^{\frac 1 3} \Rightarrow dx= \dfrac {2y} 3 \cdot {(y^2+3)^{\frac {-2} 3}}\cdot dy$

$$\int \sqrt{\dfrac {x^3-3} {x^{11}}}dx=\dfrac 2 3 \int\dfrac {y^2} {{(y^2+3)}^{\frac 5 2}}dy=\dfrac 23 \{\int \dfrac 1 {(y^2+3)^{\frac 3 2}}dy-3\int \dfrac 1 {(y^2+3)^{\frac 5 2}}dy \}= $$

$$=\dfrac 2 3 \{\dfrac {y} {3(y^2+3)^{\frac 1 2}}-\dfrac {y(2y^2+9)} {9(y^2+3)^{\frac 3 2}} \}+c=\dfrac {2y^3} {27(y^2+3)^{\frac 3 2}}+c=$$

$$=\dfrac {2(x^3-3)^{\frac 3 2}} {27x^{\frac 9 2}}+c$$

Answer 3

Let us write $$ \sqrt{\frac{x^3-3}{x^{11}}}=\frac{1}{x^4}\sqrt{1-\frac{3}{x^3}}. $$ Then, with $t=1-3/x^3$, we find that $dt=9/x^4\,dx$, and so $$ \int \sqrt{\frac{x^3-3}{x^{11}}}\,dx = \frac{1}{9}\int\sqrt{t}\,dt=\frac{1}{9}\frac{2}{3}t^{3/2}+C=\frac{2}{27}\Bigl(1-\frac{3}{x^3}\Bigr)^{3/2}+C. $$