How to integrate $\int{\frac{x^2+5}{x^3+3x}}dx$

Question

How to integrate $$\int{\frac{x^2+5}{x^3+3x}}dx$$

I have computed it in a solver with steps. It uses partial fraction decomposition. And it gets two different denominators even if there exists only one real root: $x^3+3x=x(x^2+3) \Rightarrow x = 0$. What is the most easy and most common way to solve that integral?

Answer 1

Do partial fractions:

$$\frac{x^2+5}{x(x^2+3)}=\frac Ax+\frac{Bx+C}{x^2+3}\implies x^2+5=A(x^2+3)+(Bx+C)x$$

Put now:

$$x=0\implies 5=3A\implies A=\frac53\,,\,\,\text{coeff. of}\;x^2: \; 1=A+B\implies B=-\frac23$$

and finally, comapring coefficients of $\;x\;$ in both sides wer get $\;C=0\;$, so

$$\frac{x^2+5}{x(x^2+3)}=\frac5{3x}-\frac{2x}{3(x^2+3)}$$

Now integrate the above.

Answer 2

Partial fraction decomposition :

$$\int{\frac{x^2+5}{x^3+3x}}dx=\int\frac{x^2+5}{x(x^2+3)}dx=\int\bigg(\frac{5}{3x}-\frac{2x}{3(x^2+3)}\bigg)dx = \int\frac{5}{3x}dx-\int\frac{2x}{3(x^2+3)}dx$$

$$=$$

$$\int\frac{5}{3x}dx-\int\frac{2x}{3(x^2+3)}dx = \frac{5}{3}\int \frac{1}{x}dx-\frac{2}{3}\int\frac{x}{x^2+3}$$

Substitute $u=x^2 + 3 \to dx = \frac{1}{2x}du$ which means that :

$$\int \frac{x}{x^2+3}\to\int \frac{1}{2u}du$$

Solving both the integral parts and substituting back, you'll finally get :

$$\int{\frac{x^2+5}{x^3+3x}}dx=-\dfrac{\ln\left(x^2+3\right)-5\ln\left(\left|x\right|\right)}{3}+C$$

Answer 3

HINT: I think partial fractioning may not be the best way depending on your criteria but it is a common practice:

$$\frac{x^2+5}{x^3+3x} = \frac{A}{x}+\frac{Bx+C}{x^2+3}$$

From here, you can easily see that $C=0$ and we also have $A+B = 1$ and $3A = 5$. Can you take it from here?

Answer 4

Let us try an approach where we will try to avoid explicitly finding a partial fraction decomposition.

Let $$I = \int \frac{x^2 + 5}{x^3 + 3x} \, dx.$$ Now \begin{align*} I &= \int \frac{(x^2 - 3) + 8}{x^3 + 3x} \,dx\\ &= \int \frac{x^2 - 3}{x^2 \left (x + \dfrac{3}{x} \right )} \, dx + 8 \int \frac{dx}{x^3 + 3x}\\ &= \int \frac{1 - \frac{3}{x^2}}{x + \frac{3}{x}} \, dx + 8 \int \frac{dx}{x^3 + 3x}. \end{align*} Enforcing a substitution of $u = x + 3/x$ gives $$I = \int \frac{du}{u} + \int \frac{dx}{x^3 + 3x} = \ln|u| + \int \frac{dx}{x^3 + 3x} = \ln \left |x + \frac{3}{x} \right | + 8 \int \frac{dx}{x^3 + 3x}. \tag1$$

Now consider $$2I = 2 \int \frac{x^2 + 5}{x^3 + 3x} \, dx.$$ Writing the numerator as the derivative of the denominator we have \begin{align*} 2I &= 2 \int \frac{\frac{1}{3} (3x^2 + 15)}{x^3 + 3x} \, dx\\ &= 2 \int \frac{\frac{1}{3} (3x^2 + 3) + 4}{x^3 + 3x} \, dx\\ &= \frac{2}{3} \int \frac{3x^2 + 3}{x^3 + 3x} \, dx + 8 \int \frac{dx}{x^3 + 3x}\\ &= \frac{2}{3} \ln |x^3 + 3x| + 8 \int \frac{dx}{x^3 + 3x} \tag2 \end{align*} Subtracting (2) from (1) gives $$I = \frac{2}{3} \ln |x^3 + 3x| - \ln \left |\frac{x^2 + 3}{x} \right | + C,$$ or after simplifying $$\int \frac{x^2 + 5}{x^3 + 3x} \, dx = \frac{5}{3} \ln |x| - \frac{1}{3} \ln (x^2 + 3) + C.$$

Answer 5

Using logarithmic integration, write the integral as:

$$\frac{1}{3} \int \frac{3x^2+3}{x^3+3x} \ dx + \int \frac{4}{x^3+3x} \ dx$$

and since $(x^2+3)(1) + x(-x) = 3$, $(x^2+3)(\frac{4}{3}) + x(-\frac{4}{3}x) = 4$, so this equals:

$$\frac{1}{3} \int \frac{3x^2+3}{x^3+3x} \ dx + \int \frac{4/3}{x} + \frac{(-4/3)x}{x^2+3} \ dx$$

$$= \frac{1}{3} \ln |x^3 + 3x| + \frac{4}{3} \ln |x| - \frac{2}{3} \ln|x^2+3| + C$$

which can be rewritten as:

$$\frac{1}{3} \ln |x| + \frac{1}{3} \ln |x^2 + 3| + \frac{4}{3} \ln |x| - \frac{2}{3} \ln|x^2+3| = \frac{5}{3} \ln |x| - \frac{1}{3} \ln|x^2+3|+ C.$$