How to find the integral of $\int_{o}^{u} x^{k}\frac{\alpha\theta^{\alpha}}{\left( x+\theta \right)^{\alpha+1}} dx$?

Question

I have tried to solve this by myself by letting $v=\frac{\theta}{x+θ} $... could you show me an alternative way to get the integral of $\int_{o}^{u} x^{k}\frac{\alpha\theta^{\alpha}}{\left( x+\theta \right)^{\alpha+1}} dx$?

Answer 1

We can use $x=v \theta$ substitution, $$\therefore \mathrm{d}x=\theta \mathrm{d}v$$ $$\int _0^{u/\theta} (v\theta)^k \dfrac {\alpha \theta^{\alpha +1}}{\theta^{\alpha +1}(1+v)^{\alpha + 1}}\mathrm{d}v$$ $$=\int_0^{u/\theta} \alpha\theta^k \cdot\ \frac {v^k}{(1+v)^{\alpha+1}} \mathrm{d}v$$

$$=\alpha \theta^k \int_0^{u/\theta} \dfrac {v^k}{(1+v)^{\alpha +1}} \mathrm{d}v$$

Let $I_{(k,\alpha)} = \int \dfrac {v^k}{(1+v)^{\alpha +1}} \mathrm{d}v$

Now you need to simply apply Integration by Parts on $I_{k,\theta}$ by assuming $v^k$ as the first function and $(1+v)^{\alpha +1 }$ as the second function.

After one round of IBP we see,

$$I_{(k,\alpha)} = \dfrac {-\alpha v^k}{(1+v)^{\alpha}} + \alpha k(\int \dfrac {v^{k-1}}{(1+v)^{\alpha}} \mathrm{d}v)$$

Hence we get a relation,

$$I_{(k,\alpha)} = \dfrac {v^k}{(1+v)^{\alpha +1}}+I_{(k-1,\alpha -1)}$$

Using this equation recursively we can find $I$ for any value of $k$ and $\alpha$ .