<s>[INST] $ \int \frac{x^3}{\sqrt{x^2+x}}\, dx$ I'm trying to solve this irrational integral $$ \int \frac{x^3}{\sqrt{x^2+x}}\, dx$$
doing the substitution
$$ x= \frac{t^2}{1-2 t}$$
according to the rule.
So the integral becomes:
$$  \int \frac{-2t^6}{(1-2t)^4}\, dt= \int (-\frac{1}{8}t^2-\frac{1}{4}t-\frac{5}{16}+\frac{1}{16}\frac{-80t^3+90t^2-36t+5}{(1-2t)^4})\, dt=\int (-\frac{1}{8}t^2-\frac{1}{4}t-\frac{5}{16}+\frac{1}{16}(\frac{10}{1-2t}-\frac{15}{2} \frac{1}{(1-2t)^2}+\frac{3}{(1-2t)^3}-\frac{1}{2} \frac{1}{(1-2t)^4}))\, dt=-\frac{1}{24}t^3-\frac{1}{8}t^2-\frac{5}{16}t-\frac{5}{16}\cdot \ln|1-2t|
-\frac{15}{64}\frac{1}{1-2t}+\frac{3}{64} \frac{1}{(1-2t)^2}-\frac{1}{16 \cdot 12} \frac{1}{(1-2t)^3}+cost $$
with $t=-x+ \sqrt{x^2+x}$.
The final result according to my book is instead $(\frac{1}{3}x^2-\frac{5}{12}x+\frac{15}{24})\sqrt{x^2+x}-\frac{5}{16}\ln( x+\frac{1}{2}+ \sqrt{x^2+x})$
And trying to obtain the same solution putting t in the formulas I'm definitely lost in the calculation...
I don't understant why this difference in the complexity of the solution...
Can someone show me where I'm making mistakes?
 [/INST] $$x=\frac{t^2}{1-2t}$$
$$dx=\frac{2t(1-2t)+2t^2}{(1-2t)^2}dt$$
$$=\frac{2(t-t^2)}{(1-2t)^2}dt$$
$$\sqrt{x^2+x}=x+t=\frac{t-t^2}{1-2t}$$
$$\frac{x^3dx}{\sqrt{x^2+x}}=$$
$$\frac{t^6}{(1-2t)^3}\frac{1-2t}{t-t^2}\frac{2(t-t^2)dt}{(1-2t)^2}$$
$$=\frac{2t^6dt}{(1-2t)^4}$$
To make the integration easier, put $$u=2t$$
to get
$$\frac{1}{64}\int \frac{u^6du}{(1-u)^4}$$
$$=\frac{1}{64}\int \frac{(u^6-u^5+u^5-u^4+u^4-u^3+u^3)du}{(1-u)^4}$$
Here is an other way
$$x^2+x=(x+\frac 12)^2-\frac 14$$
put
$$x+\frac 12=\frac 12\cosh(t)$$
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