Substitution Rule for Definite Integrals — Question 7

Question 7 - Solution

The integrand suggests a substitution involving $\ln x$.

Let $$u=1+\ln x.$$ Then $$du=\frac{1}{x}\,dx.$$

Change the limits of integration. When $x=1$, $u=1$. When $x=e$, $u=2$.

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

Integrate: $$\int \sqrt{u}\,du=\frac{2}{3}u^{3/2}.$$

Apply the limits: $$\frac{2}{3}u^{3/2}\Big|_{1}^{2} = \frac{2}{3}\left(2^{3/2}-1\right).$$

$$\boxed{\frac{2}{3}\left(2^{3/2}-1\right)}$$