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Prove that $\int\frac{3x^2+1}{(x^2-1)^3}dx=\frac{-x}{(x^2-1)^2}+c$

My Try:

$\int\frac{3x^2+1}{(x^2-1)^3}dx$

Put $x=\sec\theta$

$\int\frac{3\sec^2\theta+1}{\tan^6\theta}\sec\theta\tan\theta d\theta$

Converting to $\sin$ and $\cos$ we get

$=\int\frac{(3+\cos^2\theta)\cos^2\theta}{\sin^5\theta}d\theta$

I could not solve it further.

Brahmagupta
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    Do you necessarily have to prove the claim in that direction? Why not just differentiate the right hand side? (Seems to me like that would be easier) – user170231 Feb 16 '16 at 15:48
  • See http://math.stackexchange.com/questions/1658404/evaluate-int-frac1x212dx/1658449#1658449 – lab bhattacharjee Feb 16 '16 at 17:14

6 Answers6

3

Take the partial fractions of the integrand which are $\frac{1}{2(x-1)^3} -\frac{1}{2(x+1)^3}$ which can be easily integrate using substitution.

user41736
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HINT: $$ \frac{3x^2+1}{(x^2-1)^3}= \frac{3(x^2-1)+4}{(x^2-1)^3}= \frac{3}{(x^2-1)^2}+\frac{4}{(x^2-1)^3} $$

then write $(x^2-1)=(x+1)(x-1)$ and try to split the fractions.

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

juantheron
  • 53,015
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$$I = \int\frac{3x^2+1}{(x^2-1)^3}dx = \int\frac{3x^2+1}{(x-1)^3\cdot (x+1)^3}dx = \frac{1}{2}\int \left[\frac{1}{(x-1)^3}-\frac{1}{(x+1)^3}\right]dx$$

So we get $$I = \frac{1}{2}\left[-\frac{1}{2(x-1)^2}+\frac{1}{2(x+1)^2}\right]+\mathcal{C} = -\frac{1}{4}\cdot\frac{4x}{(x^2-1)^2}+\mathcal{C}$$

juantheron
  • 53,015
1

$\dfrac{d\left(\dfrac{ax+b}{(x^2-1)^n}\right)}{dx}=-\dfrac a{(x^2-1)^n}-\dfrac{2a}{(x^2-1)^{n+1}}-b\cdot\dfrac{2x}{(x^2-1)^{n+1}}$

Integrating both sides wrt $x,$

$$-\dfrac{ax+b}{(x^2-1)^n}=a\int\dfrac{dx}{(x^2-1)^n}+2a\int\dfrac{dx}{(x^2-1)^{n+1}}+b\int\dfrac{2x\ dx}{(x^2-1)^{n+1}}$$

$$\implies2a\int\dfrac{dx}{(x^2-1)^{n+1}}=-\dfrac{ax+b}{(x^2-1)^n}-2a\int\dfrac{dx}{(x^2-1)^{n+1}}+\dfrac b{n(x^2-1)^n}\ \ \ \ (1)$$

Now $\displaystyle\int\dfrac{3x^2+1}{(x^2-1)^3}dx=\int\dfrac{3(x^2-1)+4}{(x^2-1)^3}dx=3\int\dfrac{dx}{(x^2-1)^2}+4\int\dfrac{dx}{(x^2-1)^3}$

Set $n=1,2$ in $(1)$

lab bhattacharjee
  • 274,582
0

Another way is, simply, to take the derivative of $$\frac{-x}{(x^2-1)^2}+c$$ and verify that this derivative is no other that $$\frac{3x^2+1}{(x^2-1)^3}$$.

Piquito
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