Need answer to a basic indefinite integral

Question

Integrate $$\int \frac{x^2 + 3}{x^6(x^2 + 1)}\,\mathrm{d}x$$ with respect to x.

I have no idea how to approach this problem. So it'll be better if you can provide the steps...

Answer 1

Begin with some algebra: use partial fractions decomposition. To do that faster, set $u=x^2$. Then: $$\frac{u+3}{u^3(u+1)}=\frac Au+\frac B{u^2}+\frac C{u^3}+\frac D{u+1}$$ To obtain the coefficients, the faster way is to use the division algorithm along increasing powers of $u$ up to degree $3$, for $u+3$ by $u+1$. This gives: \begin{align*}3+u&=(1+u)(3-2u-2u^2)-2u^4\\ \text{whence}\quad\frac{3+u}{u^3(1+u)}&=\frac3{u^3}-\frac2{u^2}-\frac2u-\frac{2u}{1+u}\\ &=\frac3{u^3}-\frac2{u^2}-\frac2u-2+\frac{2}{1+u}. \end{align*} Now replace $u$ with $x^2$ and integrate. you should find: $$\frac{-30x^6+30x^4+10x^2-9}{15x^5}+2\arctan x.$$

Answer 2

$$\int \left(\frac{x^2+3}{x^6(x^2+1)}\right)\text{d}x=$$ $$\int \left(\frac{3}{x^6}-\frac{2}{x^4}+\frac{2}{x^2}-\frac{2}{x^2+1}\right)\text{d}x=$$ $$\int \frac{3}{x^6}\text{d}x-\int \frac{2}{x^4}\text{d}x+\int \frac{2}{x^2}\text{d}x-\int \frac{2}{x^2+1}\text{d}x=$$ $$3 \int \frac{1}{x^6}\text{d}x-2\int \frac{1}{x^4}\text{d}x+2\int \frac{1}{x^2}\text{d}x-2\int \frac{1}{x^2+1}\text{d}x=$$ $$3 \int x^{-6}\text{d}x-2\int x^{-4}\text{d}x+2\int x^{-2}\text{d}x-2\tan^{-1}(x)=$$ $$3 \cdot \frac{1}{-6+1}x^{-6+1}-2\cdot \frac{1}{-4+1}x^{-4+1}+2\cdot \frac{1}{-2+1}x^{-2+1}-2\tan^{-1}(x)+C=$$ $$3 \cdot \frac{1}{-5}x^{-5}-2\cdot \frac{1}{-3}x^{-3}+2\cdot \frac{1}{-1}x^{-1}-2\tan^{-1}(x)+C=$$ $$-\frac{3}{5x^5}+\frac{2}{3x^3}-\frac{2}{x}-2\tan^{-1}(x)+C$$

Answer 3

As stated, partial fractions is one approach to this problem. Another approach I see is trig substitution to take care of that $x^2+1$ in the denominator.

$$x=\tan t,dx=\sec^2tdt$$

$$\int\dfrac{\tan^2t+3}{\tan^6t(\sec^2t)}\sec^2tdt=\int\dfrac{\tan^2t+3}{\tan^6t}dt=\int(\cot^4t+3\cot^6t)dt$$

Now from here, just repeatedly use $\cot^2t=\csc^2t-1$, then substitute

$$u=\cot t,du=-\csc^2tdt$$