Indefinite integral with irrational exponent: $\int \frac{(2+x^{\sqrt{2}})^{3/2}}{x}dx$

Question

I cannot start to work on this $\int \frac{(2+x^{\sqrt{2}})^{3/2}}{x}dx.$

Answer 1

Let $u = 2+x^{\sqrt{2}}$. Then $du=\sqrt{2}x^{\sqrt{2}-1}dx=\sqrt{2}(u-2)\frac{dx}{x}$, so $\frac{dx}{x}=\frac{du}{\sqrt{2}(u-2)}$.

$$I=\int \frac{(2+x^{\sqrt{2}})^{3/2}}{x}dx=\frac{1}{\sqrt{2}}\int \frac{u^{3/2}}{(u-2)}du$$ Now let $v=\sqrt{u}$, so $dv=\frac{du}{2\sqrt{u}}$. Therefore

\begin{align*} I&=\sqrt{2}\int \frac{v^4}{v^2-2}dv=\sqrt{2}\int \left(\frac{4}{v^2-2}+v^2+2\right)dv\\ &=4\sqrt{2}\int \frac{1}{v^2-2}dv+\sqrt{2}\int (v^2+2)dv\\ &=-4\sqrt{2}\int \frac{1}{2\left(1-\frac{v^2}{2}\right)}dv+\sqrt{2}\frac{v^3}{3}+2\sqrt{2}v+c\\ &=-2\sqrt{2}\int \frac{1}{1-\frac{v^2}{2}}dv+\sqrt{2}\frac{v^3}{3}+2\sqrt{2}v+c\\ \end{align*}

Now for $\int \frac{1}{1-\frac{v^2}{2}}dv$ let $w=\frac{v}{\sqrt{2}}$, so $dw=\frac{dv}{\sqrt{2}}$ $$-2\sqrt{2}\int \frac{1}{1-\frac{v^2}{2}}dv=-4\int \frac{1}{1-w^2} dw=-4\tanh^{-1}(w)$$ So sub in $w=\frac{v}{\sqrt{2}}$, and then $v=\sqrt{u}$ to give \begin{align*} I&=-4\tanh^{-1}\left(\frac{v}{\sqrt{2}}\right)+\sqrt{2}\frac{v^3}{3}+2\sqrt{2}v+c\\ &=-4\tanh^{-1}\left(\frac{u^{1/2}}{\sqrt{2}}\right)+\frac{\sqrt{2}u^{3/2}}{3}+2\sqrt{2}u^{1/2}+c\\ \end{align*} Now sub in $u = 2+x^{\sqrt{2}}$ to give \begin{align*} I&=-4\tanh^{-1}\left(\frac{(2+x^{\sqrt{2}})^{1/2}}{\sqrt{2}}\right)+\frac{\sqrt{2}(2+x^{\sqrt{2}})^{3/2}}{3}+2\sqrt{2}(2+x^{\sqrt{2}})^{1/2}+c\\ &=-4\tanh^{-1}\left(\frac{(2+x^{\sqrt{2}})^{1/2}}{\sqrt{2}}\right)+\sqrt{2}(2+x^{\sqrt{2}})^{1/2}\left(\frac{8+x^{\sqrt{2}}}{3}\right)+c \end{align*}