Question
Evaluate the following integrals:$\int\frac{\text{ax}^3+\text{bx}}{\text{x}^4+\text{c}^2}\text{ dx}$

Answer

Let $\text{I}=\int\frac{\text{ax}^3+\text{bx}}{\text{x}^4+\text{c}^2}\text{ dx}$ Let $\text{ax}^3+\text{bx}=\lambda\frac{\text{d}}{\text{dx}}\big(\text{x}^4+\text{c}^2\big)+\mu$ $\text{ax}^3+\text{bx}=\lambda\big(4\text{x}^3\big)+\mu$ Comparing the coefficients of like powers of x, $4\lambda=\text{a}\Rightarrow\lambda=\frac{\text{a}}{4}$ $\mu=0\Rightarrow\mu=0$ So, $\text{I}=\int\frac{\frac{\text{a}}{4}(4\text{x}^3)+\text{bx}}{\text{x}^4+\text{c}^2}\text{ dx}$ $\text{I}=\frac{\text{a}}{4}\int\frac{4\text{x}^3}{\text{x}^4+\text{c}^2}\text{ dx}+\text{b}\int\frac{\text{x}}{(\text{x}^2)^2+\text{c}^2}\text{ dx}$ $\text{I}=\frac{\text{a}}{4}\int\frac{4\text{x}^3}{\text{x}^4+\text{c}^2}\text{ dx}+\frac{\text{b}}{2}\int\frac{2\text{x}}{(\text{x}^2)^2+\text{c}^2}\text{ dx}$ $\text{I}=\frac{\text{a}}{4}\log\big|\text{x}^4+\text{c}^2\big|+\frac{\text{b}}{2}\text{I}_1\ ....(1)$Now,
$\text{I}_1=\int\frac{2\text{x}}{(\text{x}^2)^2+\text{c}^2}\text{ dx}$ Put $\text{x}^2=\text{t}$ $\Rightarrow2\text{x}\text{ dx}=\text{dt}$ $\text{I}_1=\int\frac{1}{(\text{t})^2+\text{c}^2}\text{ dx}$ $\text{I}_1=\frac{1}{\text{c}}\tan^{-1}\Big(\frac{\text{t}}{\text{c}}\Big)+\text{C}_1\ ....(2)$ Using equation (2) in equation (1), $\text{I}=\frac{\text{a}}{4}\log\big|\text{x}^4+\text{c}^4\big|+\frac{\text{b}}{2\text{c}}\tan^{-1}\Big(\frac{\text{x}^2}{\text{c}}\Big)+\text{K}$ K = Integration constant.

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