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Can you guys help me solve this integral? $$ \int \frac{x^3}{x^4 + 2x^2 - 6}dx $$

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closed as off-topic by Watson, Shailesh, Claude Leibovici, Davide Giraudo, R_D Jul 25 '16 at 14:17

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Let $t=x^2+1$ then your integral becomes $$\frac{1}{2}\int \frac{t-1}{t^2-7}\,dt=\frac{1}{2}\int \frac{tdt}{t^2-7}- \frac{1}{2}\int \frac{dt}{t^2-7}.$$ Now $$\int \frac{tdt}{t^2-7}=\frac{1}{2}\ln(t^2-7)$$ and $$\int \frac{dt}{t^2-7}=-\frac{1}{\sqrt{7}}\mbox{arctanh}(t/\sqrt{7})$$

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HINT:

$$x^4+2x^2-6=(x^2+1)^2-7$$

Let $x^2+1=y\implies2x\ dx=dy$ and $x^2=y-1$

$$\dfrac{2Ay+B}{y^2-7}=A\cdot\dfrac{2y}{y^2-7}+\dfrac B{y^2-7}$$

Use $\dfrac{2a\ du}{u^2-a^2}=\dfrac{u+a-(u-a)}{(u+a)(u-a)}=?$

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$$I= \int \frac{x^3}{x^4 + 2x^2 - 6}dx$$

As Arthur suggests, let $u = x^2,du = 2x$

$$I = \frac{1}{2}\int \frac{u}{u^2 + 2u - 6}du$$ $$ = \frac{1}{4\sqrt{7}} (\int \frac{u}{u+1-\sqrt{7}}du -\int \frac{u}{u+1+\sqrt{7}}du ) $$

Let $z = u+1-\sqrt{7},w = u+1+\sqrt{7}$

$$I = \frac{1}{4\sqrt{7}} (\int \frac{z-1+\sqrt{7}}{z}dz -\int \frac{w-1-\sqrt{7}}{w}dw) = \frac{1}{4\sqrt{7}} (z +(\sqrt{7}-1)\ln{z} - w+(1+\sqrt{7}) \ln w)\\ = \frac{1}{4\sqrt{7}} (u+1-\sqrt{7} +(\sqrt{7}-1)\ln(u+1-\sqrt{7}) - u+1+\sqrt{7} + (1+\sqrt{7})\ln (u+1+\sqrt{7}) ) \\ = \frac{1}{4\sqrt{7}}(2+(\sqrt{7}-1)\ln(u+1-\sqrt{7})+(\sqrt{7}-1)\ln(u+1-\sqrt{7})) = \frac{1}{4\sqrt{7}}(2+(\sqrt{7}-1)\ln(x^2+1-\sqrt{7})+(\sqrt{7}-1)\ln(x^2+1-\sqrt{7}))$$

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