Question

eᵃ log x+ eˣ log a ,Integrate the given function w.r.t. x considering them well defined and integrable over proper domain.

Answer 1

Dear Student,

Solution:

$\int{e}^{a} log \: x + {e}^{x} log \: a \: dx$

after applying linearity taking 1st function

$\int {e}^{a} log \: x \: dx$
as we know that
$\int UV \: dx = U \int\: V \: dx - \int\frac{dU}{dx}\int( V \: dx)dx \\ \\ = log \: x \int\: {e}^{a} dx - \int\frac{d \: log \: x}{dx}\int ({e}^{a} dx)dx \\ \\ = x \: log \: x \: {e}^{a} - \int\frac{1}{x} x {e}^{a} dx \\ \\ = x \: log \: x \: {e}^{a} - \int{e}^{a} dx \\ \\ = x \: log \: x \: {e}^{a} - x {e}^{a} + C \\ \\ = x \: {e}^{a} (log \: x - 1) + C$

For second function

$\int {e}^{x} log \: a \: dx \\ = log \: a \int\: {e}^{x} dx - \int\frac{log \: a}{dx} \int\: ({e}^{x} dx) \: dx \\ \\ = log \: a \: {e}^{x} - \int0 . {e}^{x} dx \\ \\ = loga \: {e}^{x} + c$
$\int{e}^{a} log \: x + {e}^{x} log \: a \: dx = x \: {e}^{a} (log \: x - 1) + {e}^{x} \: log \: a + C$

Hope it helps you.