More Substitution Rule — Question 10

Question 10 - Solution

Begin with an exponential substitution. Let $$u=e^x.$$ Then $$du=e^x\,dx.$$

Substitute into the integral: $$\int \frac{e^x}{\sqrt{1+e^{2x}}}\,dx = \int \frac{1}{\sqrt{1+u^2}}\,du.$$

Inverse trigonometric step (expanded):

We recognize the integrand as the derivative of $\operatorname{arsinh} u$ or $\arctan u$–type expressions. Use a trigonometric substitution to justify the result.

Let $$u=\tan\theta.$$ Then $$du=\sec^2\theta\,d\theta, \qquad \sqrt{1+u^2}=\sqrt{1+\tan^2\theta}=\sec\theta.$$

Substitute: $$\int \frac{1}{\sqrt{1+u^2}}\,du = \int \frac{\sec^2\theta}{\sec\theta}\,d\theta = \int \sec\theta\,d\theta.$$

Recall the standard integral: $$\int \sec\theta\,d\theta = \ln|\sec\theta+\tan\theta|.$$

Substitute back using $$\tan\theta=u, \qquad \sec\theta=\sqrt{1+u^2}.$$

Thus, $$\int \frac{1}{\sqrt{1+u^2}}\,du = \ln\!\left(u+\sqrt{1+u^2}\right).$$

Substitute back $u=e^x$: $$\boxed{ \ln\!\left(e^x+\sqrt{1+e^{2x}}\right)+C }$$