Question

Evaluate the function w.r.t.x $\int [ e^{(2-5x)}+ \frac{2}{6x+1}]\ dx$

Answer 1

evaluation of the function w.r.t.x is very easy,it goes like this.

∫ [ e^{(2-5x)}+ \frac{2}{6x+1}]\ dx

Answer 2

Answer:

$\frac{e^{2x-5} }{2} + \frac{log(6x+1)}{3}$

Step-by-step explanation:

Since we know that $\int\ {e^{x} } \, dx = e^{x}$.....................(1)

$\int\ {\frac{1}{x} } \, dx =logx$..............................(2)

we need to evaluate

$\int [ e^{(2-5x)}+ \frac{2}{6x+1}]\ dx$ = $\int [ e^{(2-5x)}+ \frac{2}{6x+1}]\ dx = \int\ {e^{2-5x} } \, dx +\int\ {\frac{2}{6x+1}} \, dx$

So, using (1) and (2) we get ,

                         = $\frac{e^{2x-5} }{2} + 2\frac{log(6x+1)}{6}$

                         = $\frac{e^{2x-5} }{2} + \frac{log(6x+1)}{3}$