\int \frac{x^2 + 1}{x^4 + 1} \, dx

1. Why break the fraction?

The denominator $x^4+1$ is a quartic. Integrating a quartic directly is hard, so we factor it:

$x^4+1=(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)$

Once factored, we can try partial fractions, writing

$\frac{x^2+1}{x^4+1}=\frac{Ax+B}{x^2+\sqrt{2}x+1}+\frac{Cx+D}{x^2-\sqrt{2}x+1}$

2. Solving for the constants

By equating coefficients of like powers of $x$, we find

• $A=C=0$
• $B=D=\dfrac12$

So the integrand becomes

$\frac{x^2+1}{x^4+1}=\frac12\left[\frac{1}{x^2+\sqrt{2}x+1}+\frac{1}{x^2-\sqrt{2}x+1}\right]$

3. Completing the square

Each quadratic takes the form

$x^2\pm \sqrt{2}x+1=\left(x\pm\frac{\sqrt{2}}2\right)^2+\frac12$

Now every term is of the standard type $\displaystyle\int\frac{dx}{u^2+a^2}=\frac1a\tan^{-1}\!\left(\frac{u}{a}\right)$.

4. Integrate each part

Let

• $u_1=x+\dfrac{\sqrt{2}}2\,,\ a=\dfrac{1}{\sqrt{2}}$
• $u_2=x-\dfrac{\sqrt{2}}2\,,\ a=\dfrac{1}{\sqrt{2}}$

Then

$\int\frac{dx}{u_1^2+a^2}=\sqrt{2}\,\tan^{-1}(u_1\sqrt{2})$ $\int\frac{dx}{u_2^2+a^2}=\sqrt{2}\,\tan^{-1}(u_2\sqrt{2})$

5. Combine the results

Add the two answers and multiply by $\dfrac12$ (the factor we had pulled out):

$\boxed{\displaystyle \int\!\frac{x^2+1}{x^4+1}\,dx=\frac{\sqrt{2}}2\Big[\tan^{-1}(\sqrt{2}x+1)+\tan^{-1}(\sqrt{2}x-1)\Big]+C}$

The steps use three chief ideas that frequently appear in JEE:

1. Factorisation of quartics
2. Partial-fraction decomposition
3. Standard arctan integral $\displaystyle\int\frac{dx}{x^2+a^2}=\frac1a\tan^{-1}\!\left(\frac{x}{a}\right)$

🧒🏼 Imagine breaking a big biscuit into two easy-to-eat pieces

When you see a scary fraction like $\dfrac{x^2+1}{x^4+1}$, think of it as one big biscuit. If we split it into two smaller pieces, each piece is easy to chew (integrate).

1. Break the big biscuit: we write it as two simpler fractions that add up to the same thing.
2. Shape of the small pieces: each small piece now looks like $\dfrac{1}{(x+\text{number})^2+\text{number}}$, exactly the shape whose area we already know (it turns into an arctan).
3. Eat (integrate) each smaller piece using the arctan rule.
4. Join the answers back and add a +C (the magic constant).

Step-by-step working

1. Factor the denominator
   $x^4+1=(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)$

2. Set up partial fractions
   $\frac{x^2+1}{x^4+1}=\frac{Ax+B}{x^2+\sqrt{2}x+1}+\frac{Cx+D}{x^2-\sqrt{2}x+1}$

3. Equate coefficients (compare both sides after multiplying through): [ \begin{aligned} &A+C=0,\ &-A\sqrt{2}+B+C\sqrt{2}+D=1,\ &A+C+\sqrt{2}(-B+D)=0,\ &B+D=1. \end{aligned} ] Solving gives $A=C=0,\;B=D=\dfrac12$.

4. Rewrite integrand
   $\frac{x^2+1}{x^4+1}=\frac12\Bigg[\frac{1}{x^2+\sqrt{2}x+1}+\frac{1}{x^2-\sqrt{2}x+1}\Bigg]$

5. Complete the square
   [ x^2\pm\sqrt{2}x+1=\left(x\pm\frac{\sqrt{2}}2\right)^2+\frac12. ]

6. Integrate each term
   [ \int\frac{dx}{\left(x!\pm!\frac{\sqrt{2}}2\right)^2+\frac12}=\sqrt{2},\tan^{-1}!\Big(\sqrt{2},x!\pm!1\Big)+C. ]

7. Combine and simplify
   [ \begin{aligned} \int\frac{x^2+1}{x^4+1},dx&=\frac12\Big[\sqrt{2},\tan^{-1}(\sqrt{2}x+1)+\sqrt{2},\tan^{-1}(\sqrt{2}x-1)\Big]+C\[6pt] &=\frac{\sqrt{2}}2\Big[\tan^{-1}(\sqrt{2}x+1)+\tan^{-1}(\sqrt{2}x-1)\Big]+C. \end{aligned} ]

Final Answer
$\boxed{\displaystyle \frac{\sqrt{2}}2\Big[\tan^{-1}(\sqrt{2}x+1)+\tan^{-1}(\sqrt{2}x-1)\Big]+C}$