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

Question

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

What I did :

Method $(1) $ Re writing $x^2-2 = (x^2+2)-4 $ and partial fractions.

Method $(2) $ Substituting $x^2 = 2\tan^2 \theta $

Is there any other easy methods ?

Some substitution ?

Answer 1

If $$I_n=\int\frac{1}{(ax^2+b)^n}dx\quad,\quad n\ge 2$$ $$I_1=\frac{\sqrt{\frac{b}{a}}}{b}\tan^{-1}\left(\sqrt{\frac{a}{b}}x\right)$$ then $$I_n=\frac{2n-3}{2b(n-1)}I_{n-1}+\frac{x}{2b(n-1)(ax^2+b)^{n-1}}$$ set $a=1$ and $b=2$ apply $$\frac{x^2-2}{(x^2+2)^3}=\frac{1}{(x^2+2)^2}-\frac{4}{(x^2+2)^3}$$ we have $$\int \frac{x^2-2}{(x^2+2)^3}dx=-\left(\frac{\sqrt{2}}{16}\tan^{-1}\left(\frac{\sqrt{2}x}{2}\right)+\frac{x}{8x^2+16}+\frac{x}{2(x^2+2)^2}\right)+c$$

Answer 2

$$\int \frac { x^{ 2 }-2 }{ \left( x^{ 2 }+2 \right) ^{ 3 } } dx=\int { \frac { { x }^{ 2 }+2-4 }{ { \left( x^{ 2 }+2 \right) }^{ 3 } } } dx=\underbrace { \int { \frac { dx }{ { \left( x^{ 2 }+2 \right) }^{ 2 } } } }_{ { I }_{ 1 } } -4\underbrace { \int { \frac { dx }{ { \left( x^{ 2 }+2 \right) }^{ 3 } } } }_{ { I }_{ 2 } } =\\ x=\sqrt { 2 } \tan { t } \\ dx=\sqrt { 2 } \frac { dt }{ \cos ^{ 2 }{ t } } \\ \\ { I }_{ 1 }=\int { \frac { dx }{ { \left( x^{ 2 }+2 \right) }^{ 2 } } } =\sqrt { 2 } \int { \frac { dt }{ 4{ \left( \tan ^{ 2 }{ t } +1 \right) }^{ 2 }\cos ^{ 2 }{ t } } } =\frac { \sqrt { 2 } }{ 4 } \int { \cos ^{ 2 }{ t } dt } =\frac { \sqrt { 2 } }{ 4 } \int { \frac { 1+\cos { 2t } }{ 2 } } dt=\\ =\frac { \sqrt { 2 } }{ 8 } \left( t+\frac { \sin { 2t } }{ 2 } \right) =\frac { \sqrt { 2 } }{ 8 } \left( \arctan { \left( \frac { x }{ \sqrt { 2 } } \right) } +\frac { \sin { 2\left( \arctan { \left( \frac { x }{ \sqrt { 2 } } \right) } \right) } }{ 2 } \right) \\ \\ { I }_{ 2 }=\int { \frac { dx }{ { \left( x^{ 2 }+2 \right) }^{ 3 } } } =\sqrt { 2 } \int { \frac { dx }{ { 8\left( \tan ^{ 2 }{ t } +1 \right) }^{ 3 }\cos ^{ 2 }{ t } } } =\frac { \sqrt { 2 } }{ 8 } \int { \cos ^{ 4 }{ t } dt } =\frac { \sqrt { 2 } }{ 8 } \int { { \left( \frac { 1+\cos { 2t } }{ 2 } \right) }^{ 2 } } dt=\\ =\frac { \sqrt { 2 } }{ 32 } \int { \left( 1+2\cos { 2t+\cos ^{ 2 }{ 2t } } \right) } dt=\frac { \sqrt { 2 } }{ 32 } \int { \left( 1+2\cos { 2t } +\frac { 1+\cos { 4t } }{ 2 } \right) } dt=\\ =\frac { \sqrt { 2 } }{ 32 } \left( \frac { 3t }{ 2 } +\sin { 2t+\frac { \sin { 4t } }{ 8 } } \right) =\frac { \sqrt { 2 } }{ 32 } \left( \frac { 3\arctan { \left( \frac { x }{ \sqrt { 2 } } \right) } }{ 2 } +\sin { 2\left( \arctan { \left( \frac { x }{ \sqrt { 2 } } \right) } \right) } +\frac { \sin { 4\left( \arctan { \left( \frac { x }{ \sqrt { 2 } } \right) } \right) } }{ 8 } \right) $$ so the anwer will be $$\int \frac { x^{ 2 }-2 }{ \left( x^{ 2 }+2 \right) ^{ 3 } } dx=\frac { \sqrt { 2 } }{ 8 } \left( \frac { 3\sin { 2\left( \arctan { \left( \frac { x }{ \sqrt { 2 } } \right) } \right) } }{ 2 } -\frac { \arctan { \left( \frac { x }{ \sqrt { 2 } } \right) } }{ 2 } +\frac { \sin { 4\left( \arctan { \left( \frac { x }{ \sqrt { 2 } } \right) } \right) } }{ 8 } \right) $$ further you can simplify this "monster"

Answer 3

A standard method for dealing with a numerator that is a power of positive definite quadratic polynomial is the following.

Observe that for any positive integer $n$ we have $$ D\frac{x}{(x^2+2)^n}=-\frac{(2n-1)x^2-2}{(x^2+2)^{n+1}}.\qquad(*) $$ You can write the right hand side as a linear combination of $1/(x^2+2)^n$ and $1/(x^2+2)^{n+1}$.

Interpreting $(*)$ as an integration formula then gives you a linear equation involving $\int (x^2+2)^{-(n+1)}\,dx$, $\int (x^2+2)^{-n}\,dx$ and $x/(x^2+2)^n$.

So:

• Setting $n=2$ in the resulting equation allows you to integrate $1/(x^2+2)^3$ if you know how to integrate $1/(x^2+2)^2$.
• Setting $n=1$ in the resulting equation allows you to integrate $1/(x^2+2)^2$ if you know how to integrate $1/(x^2+2)$.
• But you know how to integrate $1/(x^2+2)$, don't you?

Answer 4

Just to follow the answer above, "..for dealing with a numerator that is a power of positive definite quadratic polynomial".

Use integration by parts, \begin{align*} I_{n}:=\int\frac{dx}{\left(x^{2}+a^{2}\right)^{n}} & =\frac{x}{\left(x^{2}+a^{2}\right)^{n}}+2n\int\frac{x^{2}}{\left(x^{2}+a^{2}\right)^{n+1}}dx\\ & =\frac{x}{\left(x^{2}+a^{2}\right)^{n}}+2n\int\frac{1}{\left(x^{2}+a^{2}\right)^{n}}dx-2na^{2}\int\frac{1}{\left(x^{2}+a^{2}\right)^{n+1}}dx\\ & =\frac{x}{\left(x^{2}+a^{2}\right)^{n}}+2nI_{n}-2na^{2}I_{n+1} \end{align*} and so \begin{align*} I_{n+1} & =\frac{1}{2na^{2}}\left(\left(2n-1\right)I_{n}+\frac{x}{\left(x^{2}+a^{2}\right)^{n}}\right),\quad n=1,2,\cdots;\\ I_{1} & =\int\frac{dx}{x^{2}+a^{2}}=\frac{1}{a}\arctan\left(\frac{x}{a}\right)+C. \end{align*} For $n=2$, \begin{align*} I_{3}=\int\frac{dx}{\left(x^{2}+a^{2}\right)^{3}} & =\frac{3}{4a^{2}}I_{2}+\frac{1}{4a^{2}}\frac{x}{\left(x^{2}+a^{2}\right)^{2}}\\ & =\frac{3}{8a^{4}}I_{1}+\frac{3}{8a^{4}}\frac{x}{\left(x^{2}+a^{2}\right)}+\frac{1}{4a^{2}}\frac{x}{\left(x^{2}+a^{2}\right)^{2}}\\ & =\frac{3}{8a^{5}}\arctan\left(\frac{x}{a}\right)+\frac{3}{8a^{4}}\frac{x}{\left(x^{2}+a^{2}\right)}+\frac{1}{4a^{2}}\frac{x}{\left(x^{2}+a^{2}\right)^{2}}+C. \end{align*}

Answer 5

Another approach:

Let:

$$I(c)=\int \frac{x^2-2}{x^2+c} dx$$

$$=\int \frac{x^2+c-c-2}{x^2+c} dx$$

The above step is can be avoided by long division, if it you see it as coming out of the blue. Anyways continuing:

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

$$=x-\frac{(c+2) \arctan (\frac{x}{\sqrt{c}})}{\sqrt{c}}+C_1$$

Whet $c$ and $C_1$ are two different things. $C$ here represents the constant gained when integrating. Now find

$$\frac{I''(c)}{2}+C_2=\frac{1}{2} \frac{\partial^2}{\partial c^2} \left(x-\frac{(c+2) \arctan (\frac{x}{\sqrt{c}})}{\sqrt{c}}+C_1 \right)+C_2 =\int \frac{x^2-2}{(x^2+c)^3} dx$$

Where $C_2$ is another constant gained when integrating. You can more specifically find:

$$\frac{I''(2)}{2}+C_2$$