Question

Integration of( e ^x +5)dx

Answer 1

Solution:

Given Integral:

$\displaystyle \rm \longrightarrow I =\int( {e}^{x} + 5) \: dx$

Can be written as:

$\displaystyle \rm \longrightarrow I =\int{e}^{x}\: dx + \int5 \: dx$

$\displaystyle \rm \longrightarrow I =\int{e}^{x}\: dx + 5\int dx$

We know that:

$\displaystyle \rm \longrightarrow \int{e}^{x}\: dx = {e}^{x} + C$

$\displaystyle \rm \longrightarrow \int dx =x + C$

Using this result, we get:

$\displaystyle \rm \longrightarrow I = {e}^{x} + 5x + C$

Therefore:

$\displaystyle \rm \longrightarrow \int( {e}^{x} + 5) \: dx= {e}^{x} + 5x + C$

★ Which is our required answer.

Additional Information:

$\boxed{\begin{array}{c|c}\bf f(x)&\bf\displaystyle\int\rm f(x)\:dx\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \rm k&\rm kx+C\\ \\ \rm sin(x)&\rm-cos(x)+C\\ \\ \rm cos(x)&\rm sin(x)+C\\ \\ \rm{sec}^{2}(x)&\rm tan(x)+C\\ \\ \rm{cosec}^{2}(x)&\rm-cot(x)+C\\ \\ \rm sec(x)\ tan(x)&\rm sec(x)+C\\ \\ \rm cosec(x)\ cot(x)&\rm-cosec(x)+C\\ \\ \rm tan(x)&\rm log(sec(x))+C\\ \\ \rm\dfrac{1}{x}&\rm log(x)+C\\ \\ \rm{e}^{x}&\rm{e}^{x}+C\\ \\ \rm x^{n},n\neq-1&\rm\dfrac{x^{n+1}}{n+1}+C\end{array}}$