Evaluate the integral : $$\int\frac{1-x}{(1+x)\sqrt{x+x^2+x^3}}\,dx$$
I tried through putting $x=\tan \theta$ as well as $x=\tan^2\theta$ .but I am unable to remove the square root. I also tride by putting $x+x^2+x^3=z^2$. But I could not proceed anyway...Please help...
Update :
putting $x=\frac{1-t}{1+t}$ , I get , $$\sqrt{x+x^2+x^3}=\sqrt{\frac{2-3t+t^2-t^3}{(1+t)^3}}$$
How you got $\sqrt{x+x^2+x^3}=\sqrt{(t^3+3)(1-t^2)}/(t+1)^2$ ?
On
$\bf{My\; Solution::}$ Given $$\displaystyle \int \frac{1-x}{(1+x)\sqrt{x+x^2+x^3}}dx = -\int\frac{(x^2-1)}{(x+1)^2\sqrt{x+x^2+x^3}}dx$$
We can Write it as $$\displaystyle -\int\frac{\left(1-\frac{1}{x^2}\right)}{\left(x+\frac{1}{x}+2\right)\sqrt{x+\frac{1}{x}+1}}dx$$
Now Let $$\displaystyle \left(x+\frac{1}{x}+1\right)=t^2\;,$$ Then $$\displaystyle \left(1-\frac{1}{x^2}\right)dx = 2tdt$$
So Integral Convert into $$\displaystyle - \int\frac{2t}{(t^2+1)\cdot t}dt = -\int\frac{2}{1+t^2}dt = -2\tan^{-1}(t)+\mathcal{C}$$
So Integral Convert into $$\displaystyle \int \frac{1-x}{(1+x)\sqrt{x+x^2+x^3}}dx =-2\tan^{-1}\left(\sqrt{x+\frac{1}{x}+1}\right)+\mathcal{C}$$
Hint 1: $t \mapsto (1-x)/(1+x)$
Hint 2: One can show that $t = (1-x)/(1+x)$ is it's own inverse. In other words $x = (1-t)/(1+t)$. Hence the derivative becomes. $\mathrm{d}x = -2t \,\mathrm{d}t/(t+1)^2$. Similarly we have $\sqrt{x^3+x^2+x} = \sqrt{(t^3+3)(1-t^2)}/(t+1)^2$ so we get a nice cancelation. Explicitly we have $$ \begin{align*} \sqrt{x^3+x^2+x} & = \sqrt{ \left(\frac{1-t}{1+t}\right)^3 + \left(\frac{1-t}{1+t}\right)^2 + \left(\frac{1-t}{1+t}\right) } \\ & = \sqrt{ \frac{-t^3+t^2-3t+3}{(1+t)^3}} = \sqrt{ \frac{1+t}{1+t}\frac{(1-t)(t^2+3)}{(1+t)^3}} = \frac{\sqrt{(1-t^2)(t^3+3)}}{(t+1)^2} \end{align*} $$
Hint 3: Use the substitution $\cos u \mapsto t$, what happens?