Calculate $\int{\frac{x^\alpha}{(x^2+a^2)^{\frac{3}{2}}}} \ \mathrm{d}x$ for $\alpha = 1,2,4,6$? What is the most straightforward way to calculate the following indefinite integral? $$\int{\frac{x^\alpha}{(x^2+a^2)^{\frac{3}{2}}}}\ \mathrm{d}x,$$
for $\alpha = 1, 2, 4,5,6$?
 A: HINT: use substitution $x=a\tan \theta$, 
$$\int \frac{x^{\alpha}}{(x^2+a^2)^{3/2}}dx=\int \frac{a^{\alpha}\tan^{\alpha}\theta}{\sec^3\theta}\sec^2\theta\ d\theta=a^{\alpha}\int \frac{\tan^{\alpha}\theta}{\sec\theta} d\theta $$
A: HINTS:


*

*When $\alpha=1$, substitute $u=x^2+\text{a}^2$ and $\text{d}u=2x\space\text{d}x$:
$$\mathcal{I}_1(\text{a},x)=\int\frac{x}{\left(x^2+\text{a}^2\right)^{\frac{3}{2}}}\space\text{d}x=\frac{1}{2}\int\frac{1}{u^{\frac{3}{2}}}\space\text{d}u=-\frac{1}{\sqrt{u}}+\text{C}=\text{C}-\frac{1}{\sqrt{x^2+\text{a}^2}}$$

*When $\alpha=2$, substitute $u=x^2+\text{a}^2$ and $\text{d}u=2x\space\text{d}x$:
$$\mathcal{I}_2(\text{a},x)=\int\frac{x^2}{\left(x^2+\text{a}^2\right)^{\frac{3}{2}}}\space\text{d}x=\int\frac{1}{\sqrt{x^2+\text{a}^2}}\space\text{d}x-\text{a}^2\int\frac{1}{\left(x^2+\text{a}^2\right)^{\frac{3}{2}}}\space\text{d}x$$


Use:


*

*Assume that $\text{a}\ne0$:
$$\int\frac{1}{\sqrt{x^2+\text{a}^2}}\space\text{d}x=\ln\left|x+\sqrt{x^2+\text{a}^2}\right|+\text{C}$$

*$$\int\frac{1}{\left(x^2+\text{a}^2\right)^{\frac{3}{2}}}\space\text{d}x=\frac{x}{\text{a}\sqrt{1+\frac{x^2}{\text{a}^2}}}+\text{C}$$


So:
$$\mathcal{I}_2(\text{a},x)=\ln\left|x+\sqrt{x^2+\text{a}^2}\right|+\frac{x}{\text{a}\sqrt{1+\frac{x^2}{\text{a}^2}}}+\text{C}$$
 3. When $\alpha=4$, substitute $x=\text{a}\tan(u)$ and $\text{d}x=\text{a}\sec^2(u)\space\text{d}u$, after that substitute $v=\sin(u)$ and $\text{d}v=\cos(u)\space\text{d}u$:
$$\mathcal{I}_4(\text{a},x)=\int\frac{x^4}{\left(x^2+\text{a}^2\right)^{\frac{3}{2}}}\space\text{d}x=\text{a}^2\int\frac{\tan^2(u)}{\sec(u)}\space\text{d}u=\text{a}^2\int\frac{v^4}{\left(v^2-1\right)^2}\space\text{d}v$$
