Question

Evaluate: $\int \frac{x^2+4x}{x^3+6x^2+5}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \left(\frac{x^2+4x}{x^3+6x^2+5}\right)dx \\ \mathrm{Let}{x}^3+6x^2+5=t \\ \Rightarrow \left(3x^2+12x\right)dx=dt \\ \Rightarrow \left(x^2+4x\right)dx=\frac{dt}{3} \\ \mathrm{Putting}{x}^3+6x^2+5=t\mathrm{and}\left(x^2+4x\right)dx=\frac{dt}{3} \\ \therefore I=\frac{1}{3}\int \frac{dt}{t} \\ =\frac{1}{3}\ln \left|t\right|+C \\ =\frac{1}{3}\ln \left|x^3+6x^2+5\right|+C \end{gathered}$$