(a) Find: ∫ (d x)/(√ x +√[3] x).

\[\begin{aligned} I & =\int \frac{d x}{\sqrt{x}+\sqrt[3]{x}} \\ & =\int \frac{6 y^{5} d y}{y^{3}+y^{2}} \quad x=y^{6} \text { so that } d x=6 y^{5} d y \\ & =6 \int \frac{y^{3}}{y+1} d y \\ & =6 \int\left[\left(y^{2}-y+1\right)-\frac{1}{y+1}\right] d y \\ & =6\left[\frac{y^{3}}{3}-\frac{y^{2}}{2}+y-\log |y+1|\right]+C \\ & =2 \sqrt{x}-3 \sqrt[3]{x}+6 x^{1 / 6}-6 \log \left(x^{1 / 6}+1\right)+C \end{aligned} \]

Let \(\displaystyle \sin x=t\), then \(\displaystyle \cos x\, d x=d t\)

\[\begin{aligned} \therefore \int_{0}^{\pi / 2} \frac{\cos x}{(1+\sin x)(4+\sin x)} d x & =\int_{0}^{1} \frac{d t}{(1+t)(4+t)} \\ & =\frac{1}{3}\left[\int_{0}^{1} \frac{1}{1+t} d t-\int_{0}^{1} \frac{1}{4+t} d t\right] \\ & =\frac{1}{3}\left[\left.\log (1+t)\right|_{0} ^{1}-\left.\log (4+t)\right|_{0} ^{1}\right] \\ & =\frac{1}{3}[\log 2-\log 5+\log 4] \\ & =\frac{1}{3} \log \frac{8}{5} \end{aligned} \]