# Science:Math Exam Resources/Courses/MATH101/April 2015/Question 04 (b)/Solution 1

Using partial fraction decomposition, we can write the integrand as

${\displaystyle {\frac {1}{x^{4}+x^{2}}}={\frac {1}{x^{2}(x^{2}+1)}}={\frac {A}{x}}+{\frac {B}{x^{2}}}+{\frac {Cx+D}{x^{2}+1}}}$

for some constants ${\displaystyle A,B,C,D}$. We find that ${\displaystyle A=0,B=1,C=0,D=-1}$, which implies that

${\displaystyle \int {\frac {1}{x^{4}+x^{2}}}\,dx=\int {\frac {1}{x^{2}}}\,dx-\int {\frac {1}{1+x^{2}}}\,dx.}$

The first integral on the right-hand side of the equation can be easily evaluated:

${\displaystyle \int {\frac {1}{x^{2}}}\,dx=-{\frac {1}{x}}+C.}$

On the other hand, to compute the second integral, we can use the substitution ${\displaystyle x=\tan u}$. Then ${\displaystyle 1+x^{2}=1+\tan ^{2}u=\sec ^{2}u}$ and ${\displaystyle dx=\sec ^{2}u\,du}$, which gives

${\displaystyle \int {\frac {1}{1+x^{2}}}\,dx=\int {\frac {1}{\sec ^{2}u}}\sec ^{2}u\,du=\int 1\,du=u+C=\arctan x+C.}$

Combining these, we obtain

${\displaystyle \int {\frac {1}{x^{4}+x^{2}}}dx=\color {blue}-{\frac {1}{x}}-\arctan x+C.}$