Partial Fraction Decomposition. Need help doing u-substitution for a particular integral to solve.

Question

$$\int\dfrac{3x^2+3x+1}{x^3+x}\mathrm dx$$

This is the work i currently have so far. Any tips would be appreciated to assist me with helping solve this equation.

Answer 1

Rewrite the numerator as $3x^2+1+3x$. Note that $(x^3+x)'=3x^2+1$. Now you're left to compute the remaining part which can be done using trig substitution.

$$\int \dfrac{3x^2+3x+1}{x^3+x}\mathrm dx=\int\dfrac{(3x^2+1)+3x}{x^3+x}\mathrm dx= \ln\mid x^3+x\mid+\int \dfrac{3}{x^2+1}\mathrm dx$$

---

$$\text{Let } \begin{bmatrix}x \\ \mathrm dx\end{bmatrix}=\begin{bmatrix}\tan\theta \\ \sec^2\theta \mathrm d\theta\end{bmatrix}$$ $$\int\dfrac{3}{x^2+1}\mathrm d\theta=3\int\dfrac{\sec^2\theta}{\sec^2\theta}\mathrm d\theta=3\theta+C=3\arctan x+C$$

---

$$\dfrac{3x^2+3x+1}{x^3+x}=\dfrac{A}{x}+\dfrac{Bx+C}{x^2+1}=\dfrac{(A+B)x^2+Cx+A}{x^3+x}$$ By comparing the coefficients, we get the following system of equations: $$\begin{array} AA+B=3 \\ C \ \ \ \ \ \ \ \ =3 \\ A \ \ \ \ \ \ \ \ =1\end{array}\implies \begin{bmatrix}A\\B\\C\end{bmatrix}=\begin{bmatrix}1\\2\\3\end{bmatrix}$$ Now plugging in the values of $A, B, C$ into the integrand. $$\int\dfrac{1}{x}\mathrm dx+\int\dfrac{2x+3}{x^2+1}\mathrm dx=\ln\mid x\mid+\int\dfrac{2x+1}{x^2+1}\mathrm dx+\int\dfrac{1}{x^2+1}\mathrm dx$$ $$=\ln\mid x\mid +\ln\mid x^2+1\mid+\arctan x+C=\ln\mid x^3+x\mid+\arctan x+C$$

---

$$\int\dfrac{3x^2+3x+1}{x^3+x}\mathrm dx=\ln\mid x^3+x\mid +\arctan x+C$$