I was wondering if this integral can be solve without wolfram and others:
$$\int \frac{2x+3}{x^2+\sqrt{1-x^2}}dx$$
Thanks.
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The result is a very large formula which is hard to write and to obtain by hand using only pen and paper. In this sense the only way is using computer algebra. – Juan Ospina Sep 18 '14 at 20:01
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This is a plan of attack, at least:
1) Separate into two terms, and to find $\displaystyle\int\frac{2x}{x^2+\sqrt{1-x^2}} dx$, let $u=x^2$ to get $\displaystyle\int\frac{1}{u+\sqrt{1-u}} du$, and then let $t=\sqrt{1-u}$ to obtain $$ 2\int\frac{t}{t^2-t-1} dt.$$
Now use partial fractions.
2) For the second term, let $x=\sin\theta$ to get $\displaystyle3\int\frac{1}{x^2+\sqrt{1-x^2}} dx=3\int\frac{\cos\theta}{\sin^{2}\theta+\cos\theta} d\theta$.
Next let $\theta=\tan\frac{t}{2}, \sin\theta=\frac{2t}{1+t^2}, \cos\theta=\frac{1-t^2}{1+t^2}, d\theta=\frac{2}{1+t^2} dt$ to get
$$\int\frac{2(1-t^2)}{4t^2+(1+t^2)(1-t^2)} dt=2\int\frac{t^2-1}{t^4-4t^2-1} dt.$$
Now use partial fractions.
user84413
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You can factor $t^4-4t^2-1$ as $(t^2-2)^2-5=(t^2-(2+\sqrt{5}))(t^2+(\sqrt{5}-2))=(t-\sqrt{2+\sqrt{5}})(t+\sqrt{2+\sqrt{5}})(t^2+(\sqrt{5}-2))$, and then write $\frac{t^2-1}{t^4-4t^2-1}$ as $\frac{A}{t-\sqrt{2+\sqrt{5}}}+\frac{B}{t+\sqrt{2+\sqrt{5}}}+\frac{Ct+D}{t^2+(\sqrt{5}-2)}$ and continue with partial fractions. – user84413 Sep 22 '14 at 16:17
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Hint: Amplify with the conjugate, and then make an appropriate trigonometric substitution.
Lucian
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@abii: After breaking up the parentheses and separating the initial integral into $4$ smaller ones, you'll be left with two rational integrands, which are trivial, and two for which an obvious trigonometric substitution is in order. – Lucian Sep 18 '14 at 20:15