Integration problem...?

Question

Integrate of "a^x dx" is "a^x/( lna) +C" But i don't undestand Why?Why "ln" is used?
Another is when we integrate "dx/(4^2+x^2)" then the result is
"1/4*tan^-1 x/4 '' but how "1/4 comes"
If these problems follow any rules ,tell me.

Answer 1

1.
We know that any variable z can be expressed like
z = e^(ln z) then
a^x = e^(ln a^x) but ln (a^x) = xln(a)
Then a^x = e^(x ln(a))
Then
∫a^x dx = ∫(e^(x ln(a))) dx

Let be u = x ln(a), so, du = ln(a) dx or du/ln(a) = dx
So, replacing xln(a) and dx by u, du

=∫(e^u )/ln(a) du
= 1/ln(a)∫e^u du
= (e^u)/ln(a) + c

But u = x ln(a)
Then
= (e^(x ln (a)))/ln (a) + c
= (e^ln(a^x)) /ln(a) +c
= a^x/ln(a) +c .

2.
By common factor of 4^2
(4^2 + x^2) = (1+ x^2/4^2)4^2 = (1+(x/4)^2) 4^2

Then, integrating:
∫1/(4^2 + x^2) dx
= 1/4^2∫1/(1+(x/4)^2) dx
Let be u = x/4 then du = dx/4 or 4du = dx
replacing u and du:

=1/4^2∫1/(1 + u^2) 4 du
= 1/4∫1/(1+u^2) du
= 1/4 arctan(u) +c
= 1/4 arctan(x/4) +c

Usually arctan is written like: tan^-1
Hope my explanation was good enough

Answer 2

(1)
Consider y = a^x

By calculation, if y = e^x then dy/dx = e^x
and also if y = e^F, then dy/dx = e^F*dF/dx. F is a function of x.

When y = a^x then the a value will have it's own base and power, so a = e^k.
So y = a^x = (e^k)x = e^kx
Now differentiate e^kx:
dy/dx = e^kx∙d/dx(kx) = e^kx(k) = k.e^kx.
But, e^kx = a^x and k = lna.
Therefore:
If y = a^x, then dy/dx = a^x lna

If integration returns back the original equation after it was differented, what must you do to a^x lna to get back to a^x ?


(2)
Given:
⌠1/(4² + x²) dx
The procedure would be difficult to explain.
Putting it simply there are standard formulas derived.
⌠1/(x² + a²) dx = 1/a tan^-1 x/a
here x implies x
and a implies 4.

Answer 3

There is a ln(a) in the antiderivative, because of the chain rule.

Recall that the derivative of e^x is e^x. More to the point,
( e^u )' = u' * e^u

Now, a^x = e^( x * ln a ).

Apply the above differentiation and you see that
(a^x )' = [ e^( x * ln a) ]'
= (x * ln a)' * e^( x * ln a)
= (ln a) * a^x

So the antiderivative of a^x is (a^x)/(ln a).

In your second question, you know the antiderivative of
1/( 1 + x^2) in terms of the arctangent.
The factor of 1/4 comes by changing the 4^2 in your question, into the 1 of the formula.

Answer 4

sorry but im confused too...