ae-x-dx-

ae-x-dx-

Question Number 24363 by gopikrishnan005@gmail.com last updated on 16/Nov/17

\int {(a e)}^{x} d x

Answered by ajfour last updated on 16/Nov/17

= \frac{{(a e)}^{x}}{1 + \ln a} + C .

Commented by gopikrishnan005@gmail.com last updated on 16/Nov/17

p l s e x p l a i n

Answered by abwayh last updated on 16/Nov/17

let u = {(ae)}^{x}

\ln u = xln (ae)

\ln u = x (\ln a + 1)

\frac{1}{u} du = (\ln a + 1) dx

dx = \frac{du}{u (\ln a + 1)}

\int {(ae)}^{x} dx = \int \frac{udu}{u (\ln a + 1)} = \int \frac{du}{(\ln a + 1)} = \frac{u}{(\ln a + 1)} + c

= \frac{{(ae)}^{x}}{(lna + 1)} + c

Commented by gopikrishnan005@gmail.com last updated on 17/Nov/17

t h a n k u s i r . . o n e d b t l o g e v a l u e 1 a h s i r

Answered by A1B1C1D1 last updated on 17/Nov/17

There integral must be done piecewtemose :

For the interval where :

\log (ae) = 0

\int {(ea)}^{x} dx = \frac{{(ea)}^{x}}{\ln (ea)} our x for \log (a + 1) = 0

The answere is :

{\begin{cases} x for \log (a + 1) = 0 \\ \frac{{(ea)}^{x}}{\ln (ea)} otherwtemose \overset{+ C}{} \end{cases}

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