What is the integral of 3^x?

Question

Is it (3^(x+1))/(x+1), because if so I can't see why.

Answer 1

let 3^x=u

then xlog3=logu

differentiating

ln3dx=(1/u)du or dx=(1/ln3)(1/u)du.......(1)

now integration of 3^x is

integ{3^xdx

=integ{u*(1/ln3)*(1/u)du

=int{(1/ln3)du

=(1/ln3)u +constant

put u=3^x

result is

(1/ln3)*3^x +c

Answer 2

That looks right to me.  

Answer 3

No, because x appears in the exponent.

The answer is 3^x / ln (3) +C

let 3^x = u

x ln (3) = ln (u)

ln (3) dx = du/u

dx = du / ln(3) u

∫ 3^x dx = ∫ u du / ln(3) u

= ∫ du / ln(3)

=u / ln (3) +C

= 3^x / ln (3) +C

Answer 4

yup. just add 1 to the exponent and divide it by the exponent plus one

[3^(x+1)] / x+1

Answer 5

It's (3^x)/Log(3) + constant if it's indefinite.

Answer 6

Maybe it would help to know the derivative rule for arbitrary bases with the variable in the exponent:

d/dn [ aⁿ ] = ln(a) · aⁿ

You're probably familiar with one special case where the base happens to be the constant e:

d/dn [ eⁿ ] = ln(e) · eⁿ = eⁿ

This is because ln(e) = 1

The above is the more general case though.

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That derivative can be proven easily:

y = aⁿ

ln(y) = ln(aⁿ)

Use the rules of log to factor out the exponent:

ln(y) = n·ln(a)

Take the derivative:

d/dx [ ln(y) ] = d/dx [ n·ln(a) ]

1/y · y' = ln(a)

y' = ln(a)·y

Replace y with what it is equal to in the original problem:

y' = ln(a)·aⁿ

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When you take the derivative of something with an arbitrary base, you *multiply* by the LN of the base.

Therefore, to take this integral, you want to *divide* by the LN of the base.

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∫ 3ⁿ dn = 3ⁿ/ln(3) + C

The integral you want is not what you provided; it is actually 3^(x)/ln(3).

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All you need is to recall the above; this doesn't really require using anything u-substitutions or anything fancy. If you really want to do that though, you can do many things, including the below:

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∫ 3ⁿ dn

Use e and ln, opposite operations:

= ∫ e^[ ln(3ⁿ) ] dn

Use the rule of logs that lets you factor out an exponent:

= ∫ e^[ n·ln(3) ] dn

Let u = n·ln(3)

Then du = ln(3)dx

And dx = du / ln(3)

Substitute and integrate:

→ ∫ e^[ u ] du / ln(3)

= e^[ u ] / ln(3) + C

= e^[ n·ln(3) ] / ln(3) + C

Factor out n from the exponent, and then cancel e and ln:

= e^[ ln(3) ]ⁿ / ln(3) + C

= 3ⁿ / ln(3) + C