Can someone help me with a Calculus problem?

Question

It's an Integration by Parts problem. I've been cracking my head for an hour an still can't do it.

Integral of x2^x dx

Please help, or give me some hints. Thanks :)

Answer 1

Let u = x.

Let dv = 2^x

Then, v = 2^x / ln(2)

By parts, ∫ u dv =

uv - ∫ v du =

x ( 2^x / ln(2) ) - ∫ ( 2^x / ln(2) ) dx =

( x2^x / ln(2) ) - ( 2^x / ln(2)^2 )

Answer 2

let u = x

and dv = 2^x dx

then it leads us to .. . .

du = dx

v = (2^x)/ln 2

and by formula

∫ u dv = u v - ∫ v du

thus

= x 2^x / ln 2 - ∫ [(2^x)/ln 2] dx

= x 2^x / ln 2 - (1/ln 2) ∫ (2^x) dx

= x 2^x / ln 2 -  (2^x)/ (ln 2)^2 + C

.. . .. .

special notes that you might be needing:

derivative of a^x = a^x * (ln a)

thus D[ 2^x ] = 2^x * (ln 2)

integral of a^x wrt x = a^x / (ln a)

thus ∫ (2^x) dx = (2^x)/(ln 2) + K

Answer 3

Integrate by parts

let u = x

du = 1

let dv = 2^x

v = 2^x/ln(2)

Then ⌡udv = uv - ⌡vdu

⌡x2^x = x2^x/ln(2) - 1/ln(2)⌡2^x

⌡x2^x = x2^x/ln(2) - 2^x/ln²(2)

⌡x2^x = [x2^xln(2) - 2^x]/ln²(2)

⌡x2^x = 2^x(xln(2) -1)/ln²(2)

Answer: 2^x(xln(2) -1)/ln²(2)

Answer 4

Very interesting question. but to long to explain here.  ..... ....

You have who functions. u and v

i.e. u = x       and          v = 2^x  

therefore  u' = 1   (u' means derivative of u or du/dx) and    v' = d/dx (2^x)  

which is v' = e^x ln 2

How ... ? see below

For second function v remember it is a "constant raised to a variable power" ... you have to keep variable down or change this form by removing constant using a method log rule.

by applying logarithmic rule

a^x = e^ln (a^x)

= e^x ln a  

or    2^x = e^ln (2^x)

=  e^x ln 2

Now you have function v =   2^x =  e^x ln 2

or  v = e^x ln 2

see the formula by parts

Integral u.v dx = uv - Int u' v dx

add values and not try to solve

Answer 5

integration by parts tells you:

Int(u dv) = uv - Int(vdu)

here, you have

Int(x 2^x dx)

so x=u and 2^x dx = dv

we need to know v so we need to know that int(2^x dx) = 2^x/ln(2)

(to get this result, set u=2^x, then ln u = x ln2 and du/u=ln2 dx...integrate and get the result above)

with this result we can int by parts:

Int(x 2^x dx)= x 2^x/ln(2)- 1/ln(2)Int((2^x) dx)

=x 2^x/ln(2) -1/ln(2)[2^x/ln(2)]

=2^x[x ln(2)-1)/ln(2)^2