# Thread: stuck on integration by parts

1. ## stuck on integration by parts

hello,

im stuck on an integration by parts problem:

$\int{s \times 2^s}$

here is what i have so far

$u = s$, $du = ds$
$dv = 2^{s}ds$, $v = \frac{2^{s+1}}{s+1}$

i know that

$\int{s \times 2^s} = uv - \int{vdu}$
$= s \times \frac{2^{s+1}}{s+1} - \int{\frac{2^{s+1}}{s+1}du}$

but im not sure how to continue...

could someone show me?

2. Originally Posted by absvalue
hello,

im stuck on an integration by parts problem:

$\int{s \times 2^s}$

here is what i have so far

$u = s$, $du = ds$
$dv = 2^{s}ds$, $v = \frac{2^{s+1}}{s+1}$

i know that

$\int{s \times 2^s} = uv - \int{vdu}$
$= s \times \frac{2^{s+1}}{s+1} - \int{\frac{2^{s+1}}{s+1}du}$

but im not sure how to continue...

could someone show me?
Be careful!! $\displaystyle\int 2^s\,ds\neq \frac{2^{s+1}}{s+1}$!! You can't apply the power rule when your base is a number and your exponent is a variable!!

However, observe that $2^s=e^{s\ln 2}$.

Can you proceed?

3. Ok, I think I understand...

Is this correct then?

$\int \!s2^s\, ds$

Note that $2^s = e^{s\ln{2}}$. Thus

$\int \!s2^s\, ds = \int \!se^{s\ln{2}}\,ds$

Let $u=s$ and $dv = e^{s\ln{2}}ds$.

Then $du = ds$ and $v = \frac{e^{s\ln{2}}}{\ln{2}}$.

Using integration by parts:

$\int \!se^{s\ln{2}}\,ds = uv - \int vdu$

$= s\frac{e^{s\ln{2}}}{\ln{2}} - \int \!\frac{e^{s\ln{2}}}{\ln{2}}\,ds$

$= s\frac{e^{s\ln{2}}}{\ln{2}} - \frac{e^{s\ln{2}}}{\ln^2{2}} + c$

$= s\frac{2^s}{\ln{2}} - \frac{2^s}{\ln^2{2}} + c$

4. Yes, that is correct.

It is also useful to learn as general rules: $\frac{d a^x}{dx}= ln(a) a^x$ and $\int a^x dx= \frac{a^x}{ln(a)}+ C$.

Notice that if a= e, ln(a)= ln(e)= 1 so that becomes the "usual" derivative and anti-derivative of $e^x$.