Question

integrate e^(3x-1) dx​

Answer 1

Answer:

$\frac{1}{3} {e}^{(3x - 1)}$

Step-by-step explanation:

$put\:\:3x - 1 = t \\ differentiating \: w.r.t \: \: x, \\ 3dx = dt \\ dx = \frac{1}{3} dt$

$\int {e}^{(3x - 1)}dx = \int {e}^{t}. \frac{1}{3} dt \\ = \frac{1}{3} \int {e}^{t} dt \\ = \frac{1}{3} {e}^{t} \\ = \frac{1}{3} {e}^{(3x - 1)}$

Answer 2

Answer:

$\boxed{\red{\displaystyle\sf \int e^{3x-1} = \dfrac{1}{3}e^{3x-1}+C}}$

Step-by-step explanation:

● We need to integrate the given function. The given function is ,

$\sf\red{\dashrightarrow} e^{(3x-1)} dx$

• Now for Integrating , substitute u = 3x - 1 . Therefore , differenciating both sides wrt x , we have ,

$\sf\red{\dashrightarrow} \dfrac{ du}{dx}=\dfrac{d}{dx}(3x-1) \\\\\sf\red{\dashrightarrow} \dfrac{ du}{dx}= 3\\\\\sf\red{\dashrightarrow} \boxed{ \sf dx =\dfrac{du}{3} }$

• Now substituting this value , we have ,

$\displaystyle\sf\red{\dashrightarrow} I =\int e^{3x-1} .dx \\\\\displaystyle\sf\red{\dashrightarrow } I = \int e^u \dfrac{du}{3} \\\\\displaystyle\sf\red{\dashrightarrow } I = \dfrac{1}{3}\int e^u . du \\\\\displaystyle\sf\red{\dashrightarrow } I = \dfrac{1}{3}\bigg( e^u + C \bigg) \\\\\displaystyle\sf\red{\dashrightarrow } \boxed{\pink{\sf I = \dfrac{1}{3}e^{3x-1}+C}}$