Calculate this integral $\int \frac{x^2}{4x^4+25}dx$.

Question

I have to evaluate the integral

$$\int \frac{x^2}{4x^4+25}\,dx.$$

I thought about parts integration.

Answer 1

Hint Apply the factorization $$a^4 + b^4 = (a^2- \sqrt{2} a b + b^2) (a^2 + \sqrt{2} a b + b^2)$$ to the denominator and apply the Method of Partial Fractions to decompose the integrand.

Answer 2

Hint: $$\int \frac { x^{ 2 } }{ 4x^{ 4 }+25 } dx=\int { \frac { x^{ 2 } }{ 4x^{ 4 }-20{ x }^{ 2 }+25-20{ x }^{ 2 } } dx } =\int { \frac { { x }^{ 2 } }{ { \left( 2{ x }^{ 2 }-5 \right) }^{ 2 }-20{ x }^{ 2 } } dx } =\int { \frac { x^{ 2 }dx }{ \left( 2x^{ 2 }-2\sqrt { 5 } x-5 \right) \left( 2x^{ 2 }+2\sqrt { 5 } x-5 \right) } } $$

Answer 3

Hint:

$4x^4+25$ factors as the product of two irreducible quadratic polynomials: $$4x^4+25=(2x^2+5)^2-20x^2=(2x^2-2\sqrt 5x+5)(2x^2+2\sqrt 5x+5,$$ whence the partial fractions decomposition: $$\frac{x^2}{4x^4+25}=\frac{Ax+B}{2x^2-2\sqrt 5x+5}+\frac{Cx+D}{2x^2+2\sqrt 5x+5}.$$ As $\dfrac{x^2}{4x^4+25}$ is an even function, we have $C=-A,\enspace B=D$. Setting $x=0$ shows $B=D=0$, and reducing to the same denominator yields $A=\dfrac{\sqrt 5}{20}$. Hence we have to compute: $$\frac{\sqrt5}{20}\int\frac{ x}{2x^2-2\sqrt 5x+5}\,\mathrm d x-\frac{\sqrt5}{20}\int\frac{x}{2x^2+2\sqrt 5x+5}\,\mathrm d x$$ Now split each integral: \begin{align*} \int\frac{ x}{2x^2-2\sqrt 5x+5}\,\mathrm d x&= \frac14\int\frac{4x-2\sqrt5}{2x^2-2\sqrt 5x+5}\,\mathrm d x+\frac12\int\frac{\mathrm d x}{4x^2-4\sqrt 5x+10} \\[1ex] &=\frac14\ln(2x^2-2\sqrt 5x+5)+\frac14\int\frac{\mathrm d(2x-\sqrt5)}{(2x-\sqrt5)^2+5} \\[1ex] &= \frac14\ln(2x^2-2\sqrt 5x+5)+\frac1{4\sqrt5}\,\arctan(2x-\sqrt5) \end{align*} Similarly, $$\int\frac{ x}{2x^2+2\sqrt 5x+5}\,\mathrm d x=\frac14\ln(2x^2+2\sqrt 5x+5)+\frac1{4\sqrt5}\,\arctan(2x+\sqrt5)$$ Thus we obtain $$\int\frac{x^2}{4x^4+25}\,\mathrm dx=\frac{\sqrt5}{80}\ln\biggl(\frac{2x^2-2\sqrt 5x+5}{2x^2+2\sqrt 5x+5}\biggr)+\frac{\sqrt5}{400}(\arctan(2x-\sqrt5)-\arctan(2x+\sqrt5)). $$

The last term may be further simplified using the formula $\;\arctan p-\arctan q \equiv\arctan\dfrac{p-q}{1+pq}\mod\pi$.