Integrate $\int\frac{x+1}{x^{4}+6x^{2}+4x}dx$

Question

This gives a cubic polynomial in denominator and nonfactorisable It looks so simple but i just am not able to solve it

Answer 1

Yet another way to approach this is to enforce the substitution $x\to 1/y$. Then, we have

$$\begin{align} \int\frac{x+1}{x^4+6x^2+4x}\,dx&=-\int\frac{y(y+1)}{4y^3+6y^2+1}\,dy\\\\ &=-\frac1{12}\int \frac{1}{4y^3+6y^2+1}\,d\left(4y^3+6y^2+1\right)\\\\ &=-\frac1{12}\log\left(4y^3+6y^2+1\right)\\\\ &=\frac14 \log x-\frac1{12}\log(x^3+6x+4)+C \end{align}$$

Answer 2

Hint :

Factorise the denominator as $x$ and $x^3 + 6x + 4$. Use partial fractions. The numerators are $\frac{1}{4}$ and $\frac{-1}{12}(3x^2 + 6)$ which lend themselves nicely for the simplification. The integral can now be easily solved.

Answer 3

$\displaystyle \frac{x+1}{x^4+6x^2+4x} = \frac{x+1}{x(x^3+6x+4)} = \frac{1}{12}\left[\frac{3}{x}-\frac{3x^2+6}{x^3+6x+4}\right]$