Question

Integrate the following with respect to x:

${\displaystyle \frac{1}{3}\cos (\frac{x}{3}-4)+\frac{7}{7x+9}+{\text{e}}^{\frac{x}{5}+3}}$

Answer 1

= ${\displaystyle \int [\frac{1}{3}\cos (\frac{x}{3}-4)+\frac{7}{7x+9}+{\text{e}}^{\frac{x}{5}+3}]\text{d}x}$

= ${\displaystyle \frac{1}{3}\int \cos (\frac{x}{3}-4) \text{d}x+7\int \frac{1}{7x+9} \text{d}x+\int {\text{e}}^{\frac{x}{5}+3} \text{d}x}$

= ${\displaystyle \frac{1}{3}\frac{\sin (\frac{x}{3}-4)}{\frac{1}{3}}+\frac{7\log |7x+9|}{7}+\frac{{\text{e}}^{\frac{x}{5}+3}}{\frac{1}{5}}+\text{c}}$

= ${\displaystyle \sin (\frac{x}{3}-4)+\log |7x+9|+5{\text{e}}^{\frac{x}{5}+3}+\text{c}}$