Use the substitution $t = \mathrm{e}^x$ to show that \(\displaystyle\int \mathrm{e}^{2x} \tan^{-1}

Question

Use the substitution $t = \mathrm{e}^x$ to show that $\displaystyle\int \mathrm{e}^{2x} \tan^{-1}(\mathrm{e}^x) \, \mathrm{d}x = \displaystyle\int t \tan^{-1}(t) \, \mathrm{d}t$.[2] Hence find the exact value of $\displaystyle\int_0^1 \mathrm{e}^{2x} \tan^{-1}(\mathrm{e}^x) \, \mathrm{d}x$.[5]

Accepted Answer

(a) $\int \mathrm e^{2x}\tan^{-1}(\mathrm e^x)\,\mathrm dx=\int t\tan^{-1}t\,\mathrm dt$, where $t=\mathrm e^x$. (b) $\frac{\mathrm e^2+1}{2}\tan^{-1}(\mathrm e)-\frac{\mathrm e}{2}-\frac\pi4+\frac12$.

2025 DHS P2 Q2

2025 DHS P2 Q2

Solution:

Solution locked