14. $int\frac{x^{2}dx}{(1 - x^{2})^{5/2}}$

Explanation:

Step1: Use trigonometric substitution

Let $x = \sin\theta$, then $dx=\cos\theta d\theta$. And $1 - x^{2}=1-\sin^{2}\theta=\cos^{2}\theta$. The integral becomes $\int\frac{\sin^{2}\theta\cos\theta d\theta}{(\cos^{2}\theta)^{\frac{5}{2}}}=\int\frac{\sin^{2}\theta\cos\theta d\theta}{\cos^{5}\theta}=\int\frac{\sin^{2}\theta}{\cos^{4}\theta}d\theta=\int\tan^{2}\theta\sec^{2}\theta d\theta$.

Step2: Use substitution again

Let $u = \tan\theta$, then $du=\sec^{2}\theta d\theta$. The integral $\int\tan^{2}\theta\sec^{2}\theta d\theta=\int u^{2}du$.

Step3: Integrate $u^{2}$

Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\int u^{2}du=\frac{u^{3}}{3}+C$.

Step4: Back - substitute

Since $u = \tan\theta$ and $x=\sin\theta$, $\tan\theta=\frac{x}{\sqrt{1 - x^{2}}}$. So the result is $\frac{1}{3}\left(\frac{x}{\sqrt{1 - x^{2}}}\right)^{3}+C=\frac{x^{3}}{3(1 - x^{2})^{\frac{3}{2}}}+C$.

Answer:

$\frac{x^{3}}{3(1 - x^{2})^{\frac{3}{2}}}+C$