How can I evaluate the following integral? $$\int(\sqrt{x}-x)(e^{\arctan\sqrt{x}})^2dx$$
I'd like the whole solution if possible. I tried using the substitution: $\sqrt{x}=t$, followed by: $2\arctan{t}=m$, to get: $$\int e^m\tan^2{\frac{m}{2}}\sec^2{\frac{m}{2}}\left[1-\tan{\frac{m}{2}}\right]dm$$ But it doesn't get me anywhere.
A complete solution will be sincerely appreciated.
On
To solve such a problem without literature and programs I would set
$\displaystyle \int (\sqrt{x}-x)e^{2\arctan\sqrt{x}}dx = 2\int (t^2-t^3)e^{2\arctan t}dt := 2p(t)e^{2\arctan t} + C$
with $\enspace x=t^2$ and knowing that $\enspace \displaystyle (\arctan t)’=\frac{1}{1+t^2}$ . $\enspace p(t)$ is a polynom.
Such methods I’ve learned in the school (means: it's nothing special).
The derivation of both sides by $\enspace t\enspace $ and multiplicating by $\enspace 1+t^2\enspace $ gives
$\displaystyle (t^2-t^3)(1+t^2)=(1+t^2)p’(t)+p(t)\enspace $ and we know now that the degree of the left side is $\enspace 5$
and it follows for $\enspace p(t)\enspace $ the degree $\enspace 4$ : $\enspace p(t):=a+bt+ct^2+dt^3+et^4$ .
Comparing the coefficients we get $\enspace \displaystyle (a;b;c;d;e)=(-\frac{1}{4};\frac{1}{2};-\frac{1}{2};\frac{1}{2};-\frac{1}{4})$
and therefore $\enspace \displaystyle p(t)=-\frac{1}{4}(1-2t+2t^2-2t^3+t^4)\enspace $ with $\enspace t=\sqrt{x}$.
Assume the integral can be written in the form $g(x) = f(x) e^{2\arctan \sqrt{x}}$ for some unknown function $f(x).$ Then $$ g'(x) = \left(f' + \frac{f}{\sqrt{x}(1+x)}\right)e^{2\arctan \sqrt{x}} $$ must be the integrand, meaning $$ f' + \frac{f}{\sqrt{x}(1+x)} = \sqrt{x}-x $$ or $$ \sqrt{x}(1+x)f' + f = x - x^{3/2} + x^2 - x^{5/2}. $$
Now find a particular solution of this ODE for $f$ via the method of undetermined coefficients with $f(x)=a_0 + a_1\sqrt{x} + a_2 x + a_3 x^{3/2} + a_4x^2.$ This results in six linear equations in the five unknown $a_i.$ The six equations are consistent though, with $a_4=-1/2,$ $a_3=1,$ $a_2=-1,$ $a_1=1,$ and $a_0=-1/2.$