A rational function integration

Question

Evaluate $$\int \frac{3x^2+1}{(x^2-1)^3}dx$$ I tried breaking the numerator in terms of the denominator but it didn't help much. I also tried a few substitutions buy most of them were useless. Please give some hints. Thanks.

Answer 1

Let $$I = \int\frac{3x^2+1}{(x^2-1)^3}dx = \int\frac{3x^2+1}{x^{\frac{3}{2}}\left(x^{\frac{3}{2}}-x^{-\frac{1}{2}}\right)^3}dx$$

So $$I = \int\frac{3x^{\frac{1}{2}}+x^{-\frac{3}{2}}}{\left(x^{\frac{3}{2}}-x^{-\frac{1}{2}}\right)^3}dx\;,$$ Now Put $\left(x^{\frac{3}{2}}-x^{-\frac{1}{2}}\right)=t\;,$ Then $\left(3x^{\frac{1}{2}}+x^{-\frac{3}{2}}\right)dx = 2dt$

So we get $$I = 2\int t^{-3}dt = -\frac{1}{t^2}+\mathcal{C} = -\frac{x}{(x^2-1)^2}+\mathcal{C}$$

Answer 2

hint: If you want to avoid trig substitution, and like to play with fraction decomposition, then you can write it like this: $\dfrac{1}{(x-1)^3(x+1)^3}=\dfrac{1}{2}\cdot \dfrac{(x+1)-(x-1)}{(x-1)^3(x+1)^3}=\dfrac{1}{2}\cdot \left(\dfrac{1}{(x-1)^3(x+1)^2}-\dfrac{1}{(x+1)^3(x-1)^2}\right)$, and repeat this "trick" again till you can see some "very easy" fraction...

Answer 3

HINT:

As $3x^2+1=3(x^2-1)+4,$

$$\dfrac{3x^2+1}{(x^2-1)^3}=3\cdot\dfrac1{(x^2-1)^2}+4\cdot\dfrac1{(x^2-1)^3}$$

For the first case, write the numerator as $$\dfrac{\{n+1-(n-1)\}^2}4$$

and $$\dfrac{\{n+1-(n-1)\}^3}8$$ for the second

Answer 4

Partial fractions will do the trick.

Try write $$\frac{(3x^2+1)}{(x^2-1)^3}=\frac{ax^2+bx+c}{(x-1)^3}+\frac{ex^2+fx+g}{(x+1)^3}.$$

Further note that $(x+1)^3-(x-1)^3$ leaves only the quadratic and constant terms as non-zero.

So $$\frac{1}{(x-1)^3}-\frac{1}{(x+1)^3}= \frac{-6x^2-2}{(x-1)^3(x+1)^3} =\frac {-1}{2} \cdot \frac{(3x^2+1)}{(x-1)^3(x+1)^3}.$$

The rest is trivial.

Answer 5

Because of the cube in denominator, you can say that $$\int \frac{3x^2+1}{(x^2-1)^3}\,dx=\frac{P_n(x)}{(x^2-1)^2}$$ where $P_n(x)$ is a polynomial fo degree $n$.

Now differentiate both sides to get $$\frac{3x^2+1}{(x^2-1)^3}=\frac{\left(x^2-1\right) P_n'(x)-4 x P_n(x)}{\left(x^2-1\right)^3}$$ So, the right hand side is of degree $n+1$; then, since the left hand side is of degree $2$, this implies $n=1$.

So, set $P_1(x)=a+bx$; replacing, you then have $$3x^2+1=-3 b x^2-4 a x-b$$ Comparing the terms of same power then gives $a=0$ and $b=-1$.