# Thread: stuck on integration by parts

1. ## stuck on integration by parts

hello,

im stuck on an integration by parts problem:

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

here is what i have so far

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

i know that

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

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:

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

here is what i have so far

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

i know that

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

but im not sure how to continue...

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

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

Can you proceed?

3. Ok, I think I understand...

Is this correct then?

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

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

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

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

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

Using integration by parts:

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

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

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

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

4. Yes, that is correct.

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

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