What is the integration of a^(ln(x))?

Answer 1

ln (a^(ln x)) = (ln x) (ln a)
So a^(ln x) = x^(ln a)
and therefore ∫ a^(ln x) dx
= ∫ x^(ln a) dx
= x^(ln a + 1) / (ln a + 1) + c.

which we can also write as
x . x^(ln a) / (ln a + 1) + c
= x a^(ln x) / (ln a + 1) + c.

Answer 2

Integrate a^(ln x) with respect to x.

First rearrange the terms to make it easier to integrate.

y = a^(ln x)
ln y = ln{a^(ln x)} = ln(x)*ln(a) = ln{x^(ln a)}
y = x^(ln a)

Therefore
a^(ln x) = x^(ln a)

Now we can integrate.

∫a^(ln x)dx = ∫x^(ln a)dx = x^((ln a) + 1)/((ln a) + 1) + C
= (x)x^(ln a)/((ln a) + 1) + C
= xa^(ln x)/((ln a) + 1) + C

Answer 3

y=a^(lnx) = x^ln(a); Y=∫x^ln(a)dx= x^(ln(a)+1) /(ln(a)+1)+C, unless a=1/e;
if a=1/e then Y=lnx +C;

Answer 4

(x a^In(x)) / (1+In(a))

Answer 5

Int(a^(lnx)dx)=
=Int(x'*a^(lnx)dx)=
=x*a^(lnx)-Int(x*(a^lnx)'dx)=
=x*a^(lnx)-Int(x*(a^lnx)*lna*1/xdx)=
=x*a^(lnx)-lna*Int((a^lnx)dx)

so,
(1+lna)Int((a^lnx)dx)=x*a^(lnx)
Int((a^lnx)dx)=
=(x^a^(lnx))/(1+lna)+C