Question

# Use the Table of Integrals to evaluate the integral. (Use C for the constant of integration.) \int37e^{74x}\arctan(e^{37x})dx Inverse Trigonometric Forms (92): \int u\tan^{-1}u\ du=\frac{u^{2}+1}{2}\tan^{-1}u-\frac{u}{2}+C

Integrals
Use the Table of Integrals to evaluate the integral. (Use C for the constant of integration.)
$$\int37e^{74x}\arctan(e^{37x})dx$$
Inverse Trigonometric Forms (92): $$\int u\tan^{-1}u\ du=\frac{u^{2}+1}{2}\tan^{-1}u-\frac{u}{2}+C$$

2021-05-15

Step 1
$$\int(37)\cdot e^{74x}\cdot\tan^{-1}(e^{37x})dx$$
$$\Rightarrow e^{37x}=t\Rightarrow(37)e^{37x}\cdot dx=dt$$$$=\int t\cdot\tan^{-1}(t)dt$$
From table given formula number 92:
$$\int u\cdot\tan^{-1}u\cdot du=\frac{u^{2}+1}{2}\tan^{-1}u-\frac{u}{2}+C$$
$$\Rightarrow\frac{t^{2}+1}{2}\tan^{-1}t-\frac{t}{2}+C$$
$$\Rightarrow\frac{e^{74x}+1}{2}\tan^{-1}(e^{37x})-\frac{e^{37x}}{2}+C$$