5

i'm required to evaluate this integral. I've tried factorizing but it doesn't lead me to anywhere.

$$\int\frac{x^2-1}{(x^4+3 x^2+1) \tan^{-1}\left(\frac{x^2+1}{x}\right)}\,dx$$

I've also tried letting $u = \frac{x^2+1}{x}$, $du/dx$ gets me $1-\frac{1}{x^2}$ but it doesn't seem to be working either.

Hope to receive some advise/ solutions on how to start tackling the question

Eric Wofsey
  • 330,363
jaclynx
  • 121
  • $u = \dfrac 1 x,\qquad$ $dx = \dfrac{-du}{u^2},\qquad$ $\dfrac{x^2+1}x = x + \dfrac 1 x = \dfrac 1 u + u = \dfrac{u^2+1} u,$ $$ \int\frac{x^2-1}{(x^4+3 x^2+1) \arctan\left(\frac{x^2+1}{x}\right)},dx = \int\frac{\frac 1 {u^2} - 1}{\left( \frac 1 {u^4} + \frac 3 {u^2} + 1 \right) \arctan\left( \frac{u^2+1} u \right)} , \left( \frac{-du}{u^2} \right) $$ $$ = \int \frac{1-u^2}{(1 + 3u^2 + u^4) \arctan\left( \frac{u^2+1} u \right)} , (-du) $$ This substitution results in no change in this integral at all. So I wonder if that symmetry can be exploited. $\qquad$ – Michael Hardy Nov 10 '16 at 04:30

4 Answers4

11

An useful identity to remember is $$\frac{dx}{x} = \frac{d(x+x^{-1})}{x-x^{-1}} = \frac{d(x-x^{-1})}{x + x^{-1}}$$

Using the first part of this identity, you can rewrite the integral as

$$\begin{align} & \int \frac{x(x^2-1)}{(x^4+3x^2+1)\tan^{-1}\left(x + x^{-1}\right)}\frac{d(x+x^{-1})}{x-x^{-1}}\\ = & \int \frac{x^2 d(x+x^{-1})}{(x^4+3x^2+1)\tan^{-1}\left(x + x^{-1}\right)}\\ = & \int \frac{d(x+x^{-1})}{(x^2+x^{-2}+3)\tan^{-1}\left(x + x^{-1}\right)}\\ = & \int \frac{d(x+x^{-1})}{((x+x^{-1})^2+1)\tan^{-1}\left(x + x^{-1}\right)}\\ = & \int \frac{d\tan^{-1}\left(x + x^{-1}\right)}{\tan^{-1}\left(x + x^{-1}\right)}\\ = &\log\tan^{-1}\left(x + x^{-1}\right) + \text{constant}. \end{align} $$

achille hui
  • 122,701
4

Notice, substitute $\text{u}=\text{f}\left(\text{x}\right)$ and $\text{d}\text{u}=\text{f}\space'\left(\text{x}\right)\space\text{d}\text{x}$:

$$\int\frac{\text{f}\space'\left(\text{x}\right)}{\text{f}\left(\text{x}\right)}\space\text{d}\text{x}=\int\frac{1}{\text{u}}\space\text{d}\text{u}=\ln\left|\text{u}\right|+\text{C}=\ln\left|\text{f}\left(\text{x}\right)\right|+\text{C}$$

Now, when:

$$\text{f}\left(\text{x}\right)=\arctan\left\{x+\frac{1}{x}\right\}$$

And:

$$\text{f}\space'\left(\text{x}\right)=\frac{x^2-1}{x^4+3x^2+1}$$

So, when you want to prove the result:

$$\int\frac{\frac{x^2-1}{x^4+3x^2+1}}{\arctan\left\{x+\frac{1}{x}\right\}}\space\text{d}\text{x}=\int\frac{x^2-1}{\left(x^4+3x^2+1\right)\arctan\left\{x+\frac{1}{x}\right\}}\space\text{d}\text{x}=\ln\left|\arctan\left\{x+\frac{1}{x}\right\}\right|+\text{C}$$

Jan Eerland
  • 28,671
0

Hint: use $u = \arctan(\frac {x^2 + 1}{x})$

Michael Hardy
  • 1
Horse
  • 676
0

Just my path of thinking $$ \begin{aligned} \int \frac{x^2-1}{\left(x^4+3 x^2+1\right) \tan ^{-1}\left(\frac{x^2+1}{x}\right)} d x = & \int \frac{1-\frac{1}{x^2}}{\left(x^2+\frac{1}{x^2}+3 \right)\tan ^{-1}\left(x+\frac{1}{x}\right)} d x \\ = & \int \frac{d\left(x+\frac{1}{x}\right)}{\left[\left(x+\frac{1}{x}\right)^2+1\right] \tan ^{-1}\left(x+\frac{1}{x}\right)} \\ = & \int \frac{d\left(\tan ^{-1}\left(x+\frac{1}{x}\right)\right)}{\tan ^{-1}\left(x+\frac{1}{x}\right)} \\ = & \ln \left|\tan ^{-1}\left(x+\frac{1}{x}\right)\right|+C \end{aligned} $$

Lai
  • 20,421