Evaluate integral $\int_1^\infty\frac{x^2-2}{x^4+3}dx$

Question

This is a problems that has stuck with me for days. I tried subtitution but got no where. The usual method I used seem further complicate the problem. Is there a method for this particular type of integral ? $$\int_1^\infty\frac{x^2-2}{x^4+3}dx$$ Thanks

Answer 1

Mathematica yields:

$$\frac{2 \left(\sqrt{3}-2\right) \pi +\log (144)-4 \log \left(3+\sqrt{3}-\sqrt{2} 3^{3/4}\right)+\sqrt{3} \log \left(1+\sqrt{3}+\sqrt{3+2 \sqrt{3}}\right)-2 \left(\sqrt{3}-2\right) \tan ^{-1}\left(\sqrt{3+2 \sqrt{3}}\right)}{4 \sqrt{2} 3^{3/4}}$$

which suggests it will be quite a bit of hand-calculation.

Answer 2

I will tell you the indefinite integral of this, and then you can put limits.

In general, let $a$ be any real number. Then, $$\int\frac{x^2}{x^4+a^4}dx=\frac{1}{2}\int\frac{x^2+a^2}{x^4+a^4}dx+\frac{1}{2}\int\frac{x^2-a^2}{x^4+a^4}dx$$ $$=\frac{1}{2}\int\frac{1+\frac{a^2}{x^2}}{x^2+\frac{a^4}{x^2}}dx+\frac{1}{2}\int\frac{1-\frac{a^2}{x^2}}{x^2+\frac{a^4}{x^2}}dx$$ Now apply substitution $t=x-\frac{a^2}{x}$ and $u=x+\frac{a^2}{x}$ in the above integrals to get $$\frac{1}{2}\int\frac{1}{t^2+2a^2}dt+\frac{1}{2}\int\frac{1}{u^2-2a^2}du$$ These can be solved directly now.

Similarly, $$\int\frac{1}{x^4+a^4}dx=\frac{1}{2a^2}\int\frac{x^2+a^2}{x^4+a^4}dx-\frac{1}{2a^2}\int\frac{x^2-a^2}{x^4+a^4}dx$$ Now, proceed similarly.

Hope it helps:)

Answer 3

$$I=\int_{1}^{\infty} \frac{x^2-2}{x^4+3} dx =\int_{1}^{\infty} \frac{1-2/x^2}{x^2+3x^{-2}}=\int_{1}^{\infty} \frac{A(1-\sqrt{3}/x^2)+B(1+\sqrt{3}/x^2)}{x^2+3x^{-2}} dx$$ $$\implies I=\int_{1}^{\infty}\left ( A \frac{(1-\sqrt{3}/x^2)}{(x+\sqrt{3}/x^2)-2\sqrt{3}} + B\frac{(1+\sqrt{3}/x^2)}{(x-\sqrt{3}/x^2)+2\sqrt{3}}\right) d. $$ Let $x+\sqrt{3}/x=u$ and $x-\sqrt{3}/x=v$ in the above, then $$I=\int_{1}^{\infty} \left( \frac{A du}{u^2-2\sqrt{3}}+ \frac{B dv}{v^2+2\sqrt{3}} \right), A =\frac{1}{2}(1-2/\sqrt{3}), B=\frac{1}{2}(1-2/\sqrt{3}).$$ Next, $u,v$ integrals simply do-able.

Answer 4

Rearrange as follows,

$$I=\int_1^\infty\frac{x^2-2}{x^4+3}dx =\int_1^\infty\frac{1-\frac2{x^2}}{x^2+\frac3{x^2}}dx =p_-q\int_1^\infty\frac{1+\frac{\sqrt3}{x^2}}{x^2+\frac3{x^2}}dx +p_+q\int_1^\infty\frac{1-\frac{\sqrt3}{x^2}}{x^2+\frac3{x^2}}dx$$

$$=p_-\int_1^\infty \frac{q\>d(x-\frac{\sqrt3}{x})}{(x-\frac{\sqrt3}{x})^2+q^2} +p_+\int_1^\infty \frac{q\>d(x+\frac{\sqrt3}{x})}{(x+\frac{\sqrt3}{x})^2-q^2}$$

where $p_{\pm}=\frac{\sqrt3\pm2}{(2\sqrt3)^{3/2}}$ and $q=\sqrt{2\sqrt3}$. Recognize $(\tan^{-1}t)'=\frac1{t^2+1}$ and $(\coth^{-1}t)'=-\frac1{t^2-1}$ to take the limits,

$$I=p_-\tan^{-1} \frac{x-\frac{\sqrt3}{x}}{q}\bigg|_1^\infty -p_+\coth^{-1} \frac{x+\frac{\sqrt3}{x}}{q}\bigg|_1^\infty$$

$$=p_-\left( \frac\pi2 - \tan^{-1}\frac{1-\sqrt3}{q}\right) +p_+\coth^{-1}\frac{1+\sqrt3}{q}$$

Answer 5

Consider the general problem of $$I=\int \frac{x^2+a}{x^4+b}\,dx\qquad \text{where} \qquad b >0$$ Using partial fraction decomposition $$ \frac{x^2+a}{x^4+b}=\frac i{2\sqrt b}\left(\frac{ a-i \sqrt b}{ x^2+i \sqrt b}-\frac{ a+i \sqrt b}{ x^2-i \sqrt b}\right)$$ which makes the antiderivative quite simple.