How to evaluate $\int \frac{x dx}{x^4 + 6x^2 + 5}$?

Question

$$\int \frac{x dx}{x^4 + 6x^2 + 5}$$

How to evaluate this integral ?

Answer 1

$$\int \frac { xdx }{ x^{ 4 }+6x^{ 2 }+5 } =\frac { 1 }{ 2 } \int { \frac { d{ x }^{ 2 } }{ { \left( { x }^{ 2 }+3 \right) }^{ 2 }-4 } } =\frac { 1 }{ 2 } \left[ \int { \frac { d{ x }^{ 2 } }{ \left( { x }^{ 2 }+1 \right) \left( { x }^{ 2 }+5 \right) } } \right] =\frac { 1 }{ 8 } \left[ \int { \frac { d{ x }^{ 2 } }{ \left( { x }^{ 2 }+1 \right) } -\int { \frac { d{ x }^{ 2 } }{ \left( { x }^{ 2 }+5 \right) } } } \right] =\\ =\frac { 1 }{ 8 } \left[ \int { \frac { d\left( { x }^{ 2 }+1 \right) }{ \left( { x }^{ 2 }+1 \right) } -\int { \frac { d\left( { x }^{ 2 }+5 \right) }{ \left( { x }^{ 2 }+5 \right) } } } \right] =\frac { 1 }{ 8 } \ln { \left| \frac { { x }^{ 2 }+1 }{ { x }^{ 2 }+5 } \right| } +C$$

Answer 2

Hint: $$x^4+6x^2+5=(x^2+1)(x^2+5)$$ And then make $u=x^2$.