Evaluate : \(\int \frac{5e^x}{(e^x + 1)(e^{2x} + 9)}\,dx\)

Answer: \(\frac{1}{2}\log(e^x + 1) - \frac{1}{4}\log(e^{2x} + 9) + \frac{1}{6}\tan^{-1}\left(\frac{e^x}{3}\right) + c\)

Put \(e^x = t\), so \(e^x\,dx = dt\). The integral becomes \(I = \int \frac{5}{(t + 1)(t^2 + 9)}\,dt\).

Let \(\frac{5}{(t + 1)(t^2 + 9)} = \frac{A}{t + 1} + \frac{Bt + C}{t^2 + 9}\), so \(5 = A(t^2 + 9) + (Bt + C)(t + 1)\).

Put \(t = -1\): \(5 = 10A \Rightarrow A = \frac{1}{2}\).

Coefficient of \(t^2\): \(0 = A + B \Rightarrow B = -\frac{1}{2}\). Constant term: \(5 = 9A + C \Rightarrow C = 5 - \frac{9}{2} = \frac{1}{2}\).

\(I = \frac{1}{2}\int \frac{dt}{t + 1} - \frac{1}{2}\int \frac{t}{t^2 + 9}dt + \frac{1}{2}\int \frac{dt}{t^2 + 9}\)

\(= \frac{1}{2}\log|t + 1| - \frac{1}{4}\log(t^2 + 9) + \frac{1}{2} \cdot \frac{1}{3}\tan^{-1}\left(\frac{t}{3}\right) + c\).

Resubstituting \(t = e^x\): \(I = \frac{1}{2}\log(e^x + 1) - \frac{1}{4}\log(e^{2x} + 9) + \frac{1}{6}\tan^{-1}\left(\frac{e^x}{3}\right) + c\).