$ \int \frac{x^3}{\sqrt{x^2+x}}\, dx$

Question

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?

Answer 1

Just for your curiosity since you already received @hamam_Abdallah's answer.

You can compute this integral musch faster with a different change of variable.

Because of the denominator, let $x=\sinh^2(t)$ $$\int \frac{x^3}{\sqrt{x^2+x}}\, dx=2 \int \sinh ^6(t)\,dt$$

Now, using the multiple angle formulae $$\sinh ^6(t)=\frac{15}{32} \cosh (2 t)-\frac{3}{16} \cosh (4 t)+\frac{1}{32} \cosh (6 t)-\frac{5}{16}$$ and everything becomes simple.

Answer 2

$$x=\frac{t^2}{1-2t}$$

$$dx=\frac{2t(1-2t)+2t^2}{(1-2t)^2}dt$$ $$=\frac{2(t-t^2)}{(1-2t)^2}dt$$

$$\sqrt{x^2+x}=x+t=\frac{t-t^2}{1-2t}$$

$$\frac{x^3dx}{\sqrt{x^2+x}}=$$ $$\frac{t^6}{(1-2t)^3}\frac{1-2t}{t-t^2}\frac{2(t-t^2)dt}{(1-2t)^2}$$ $$=\frac{2t^6dt}{(1-2t)^4}$$

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 $$x+\frac 12=\frac 12\cosh(t)$$

Answer 3

Prior to any substitution, apply the reduction formula $$\int \frac{x^n}{\sqrt{x^2+x}}\, dx=I_n=\frac{x^{n-1}}n \sqrt{x^2+x} -\frac{2n-1}{2n}I_{n-1} $$ (obtained via IBP) successively to simplify $$\int \frac{x^3}{\sqrt{x^2+x}}\, dx= \left(\frac{x^2}3 -\frac{5x}{12}+\frac58\right) \sqrt{x^2+x} -\frac5{16}I_0 $$ where $I_0=\int \frac{1}{\sqrt{x^2+x}}\, dx= \tanh^{-1}\frac{\sqrt{x^2+x}}{x+\frac12} $.