Question

$\int \frac{1}{x^4+1}dx$

Answer 1

$\frac{1}{4\sqrt{2}} \ln\left(\frac{x^2 + \sqrt{2}x + 1}{x^2 - \sqrt{2}x + 1}\right) + \frac{1}{2\sqrt{2}} \arctan(\sqrt{2}x+1) + \frac{1}{2\sqrt{2}} \arctan(\sqrt{2}x-1) + C$

Answer 2

To integrate $\int \frac{1}{x^4+1}dx$, we can use the technique of partial fraction decomposition.

First, factor the denominator $x^4+1$: $x^4+1 = (x^2+1)^2 - (x\sqrt{2})^2 = (x^2 - x\sqrt{2} + 1)(x^2 + x\sqrt{2} + 1)$.

Now, decompose the integrand into partial fractions: $\frac{1}{x^4+1} = \frac{Ax+B}{x^2 - \sqrt{2}x + 1} + \frac{Cx+D}{x^2 + \sqrt{2}x + 1}$

Multiplying both sides by $(x^2 - \sqrt{2}x + 1)(x^2 + \sqrt{2}x + 1)$, we get: $1 = (Ax+B)(x^2 + \sqrt{2}x + 1) + (Cx+D)(x^2 - \sqrt{2}x + 1)$

Expand and collect terms by powers of $x$: $1 = (A+C)x^3 + (\sqrt{2}A+B-\sqrt{2}C+D)x^2 + (A+\sqrt{2}B+C-\sqrt{2}D)x + (B+D)$

Comparing coefficients:

1. Coefficient of $x^3$: $A+C = 0 \implies C = -A$
2. Coefficient of $x^2$: $\sqrt{2}A+B-\sqrt{2}C+D = 0$
3. Coefficient of $x$: $A+\sqrt{2}B+C-\sqrt{2}D = 0$
4. Constant term: $B+D = 1$

From (1) and (3): $A+\sqrt{2}B-A-\sqrt{2}D = 0 \implies \sqrt{2}B-\sqrt{2}D = 0 \implies B=D$. Substitute $B=D$ into (4): $B+B=1 \implies 2B=1 \implies B=1/2$. So $D=1/2$. Substitute $C=-A$, $B=1/2$, $D=1/2$ into (2): $\sqrt{2}A+1/2-\sqrt{2}(-A)+1/2 = 0 \implies 2\sqrt{2}A+1 = 0 \implies A = -\frac{1}{2\sqrt{2}} = -\frac{\sqrt{2}}{4}$. Since $C=-A$, $C = \frac{\sqrt{2}}{4}$.

So the decomposition is: $\frac{1}{x^4+1} = \frac{-\frac{\sqrt{2}}{4}x + \frac{1}{2}}{x^2 - \sqrt{2}x + 1} + \frac{\frac{\sqrt{2}}{4}x + \frac{1}{2}}{x^2 + \sqrt{2}x + 1}$ $= \frac{1}{4} \left( \frac{-\sqrt{2}x + 2}{x^2 - \sqrt{2}x + 1} + \frac{\sqrt{2}x + 2}{x^2 + \sqrt{2}x + 1} \right)$

Now, we integrate each term. For the first term, let $I_1 = \int \frac{-\sqrt{2}x+2}{x^2 - \sqrt{2}x + 1} dx$. The derivative of the denominator is $2x-\sqrt{2}$. We rewrite the numerator: $-\sqrt{2}x+2 = -\frac{1}{\sqrt{2}}(2x-\sqrt{2}) + 1$. $I_1 = \int \left( -\frac{1}{\sqrt{2}} \frac{2x-\sqrt{2}}{x^2 - \sqrt{2}x + 1} + \frac{1}{x^2 - \sqrt{2}x + 1} \right) dx$ $I_1 = -\frac{1}{\sqrt{2}} \ln(x^2 - \sqrt{2}x + 1) + \int \frac{1}{(x - \frac{\sqrt{2}}{2})^2 + (\frac{1}{\sqrt{2}})^2} dx$ $I_1 = -\frac{1}{\sqrt{2}} \ln(x^2 - \sqrt{2}x + 1) + \frac{1}{1/\sqrt{2}} \arctan\left(\frac{x - \frac{\sqrt{2}}{2}}{1/\sqrt{2}}\right)$ $I_1 = -\frac{1}{\sqrt{2}} \ln(x^2 - \sqrt{2}x + 1) + \sqrt{2} \arctan(\sqrt{2}x - 1)$

For the second term, let $I_2 = \int \frac{\sqrt{2}x+2}{x^2 + \sqrt{2}x + 1} dx$. The derivative of the denominator is $2x+\sqrt{2}$. We rewrite the numerator: $\sqrt{2}x+2 = \frac{1}{\sqrt{2}}(2x+\sqrt{2}) + 1$. $I_2 = \int \left( \frac{1}{\sqrt{2}} \frac{2x+\sqrt{2}}{x^2 + \sqrt{2}x + 1} + \frac{1}{x^2 + \sqrt{2}x + 1} \right) dx$ $I_2 = \frac{1}{\sqrt{2}} \ln(x^2 + \sqrt{2}x + 1) + \int \frac{1}{(x + \frac{\sqrt{2}}{2})^2 + (\frac{1}{\sqrt{2}})^2} dx$ $I_2 = \frac{1}{\sqrt{2}} \ln(x^2 + \sqrt{2}x + 1) + \frac{1}{1/\sqrt{2}} \arctan\left(\frac{x + \frac{\sqrt{2}}{2}}{1/\sqrt{2}}\right)$ $I_2 = \frac{1}{\sqrt{2}} \ln(x^2 + \sqrt{2}x + 1) + \sqrt{2} \arctan(\sqrt{2}x + 1)$

Combining $I_1$ and $I_2$: $\int \frac{1}{x^4+1}dx = \frac{1}{4}(I_1 + I_2)$ $= \frac{1}{4} \left[ -\frac{1}{\sqrt{2}} \ln(x^2 - \sqrt{2}x + 1) + \sqrt{2} \arctan(\sqrt{2}x - 1) + \frac{1}{\sqrt{2}} \ln(x^2 + \sqrt{2}x + 1) + \sqrt{2} \arctan(\sqrt{2}x + 1) \right] + C$ $= \frac{1}{4} \left[ \frac{1}{\sqrt{2}} \left( \ln(x^2 + \sqrt{2}x + 1) - \ln(x^2 - \sqrt{2}x + 1) \right) + \sqrt{2} \left( \arctan(\sqrt{2}x + 1) + \arctan(\sqrt{2}x - 1) \right) \right] + C$ $= \frac{1}{4\sqrt{2}} \ln\left(\frac{x^2 + \sqrt{2}x + 1}{x^2 - \sqrt{2}x + 1}\right) + \frac{\sqrt{2}}{4} \left( \arctan(\sqrt{2}x + 1) + \arctan(\sqrt{2}x - 1) \right) + C$

This can also be written as: $= \frac{1}{4\sqrt{2}} \ln\left(\frac{x^2 + \sqrt{2}x + 1}{x^2 - \sqrt{2}x + 1}\right) + \frac{1}{2\sqrt{2}} \arctan\left(\frac{\sqrt{2}x}{1-x^2}\right) + C'$ (using $\arctan A + \arctan B = \arctan\left(\frac{A+B}{1-AB}\right)$)

Or, using the alternative decomposition method: $\int \frac{1}{x^4+1}dx = \frac{1}{2\sqrt{2}} \arctan\left(\frac{x^2-1}{\sqrt{2}x}\right) + \frac{1}{4\sqrt{2}} \ln\left(\frac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1}\right) + C$. Both forms are equivalent.

Explanation of the solution:

1. Factor the denominator: The denominator $x^4+1$ is factored into two irreducible quadratic factors: $(x^2 - \sqrt{2}x + 1)$ and $(x^2 + \sqrt{2}x + 1)$.

2. Partial Fraction Decomposition: The integrand $\frac{1}{x^4+1}$ is decomposed into partial fractions of the form $\frac{Ax+B}{x^2 - \sqrt{2}x + 1} + \frac{Cx+D}{x^2 + \sqrt{2}x + 1}$. The coefficients $A, B, C, D$ are found by comparing coefficients after cross-multiplication.

3. Integration of terms: Each resulting fraction is integrated separately.

   • For terms of the form $\int \frac{px+q}{ax^2+bx+c}dx$, the numerator is split into two parts: one proportional to the derivative of the denominator ($2ax+b$) and the other a constant.
   • The first part integrates to a logarithmic term: $\int \frac{k(2ax+b)}{ax^2+bx+c}dx = k \ln|ax^2+bx+c|$.
   • The second part integrates to an arctangent term: $\int \frac{m}{ax^2+bx+c}dx$. The denominator is completed to a square $a(x+b/2a)^2 + (c - b^2/4a)$ and then integrated using the formula $\int \frac{1}{u^2+k^2}du = \frac{1}{k}\arctan(\frac{u}{k})$.

4. Combine results: The results from integrating each partial fraction are combined to get the final integral.