How do you integrate #int x^2e^(4x)# by integration by parts method?

Question

Luke Alvarez · Accepted Answer

There is a faster method

Introduce #t# and write #I = int e^(4tx) dx# Now differentiate with respect to t #(dI)/dt = int 4x e^(4tx) dx# Once more #(d^2I)/dt^2 = int 16x^2 e^(4tx) dx# #I = e^(4tx)/(4t)# #(dI)/dt= x/te^(4tx) - e^(4tx)/(4t^2)# and again #(d^2 I)/(dt^2) = ( 4x^2/t -x/(t^2)-x/t^2 + 2/t^3 1/4)e^(4tx)# #(d^2I)/dt^2 = (4x^2-2x + 2)e^(4x)# Our integral is 1/16 I #J= I/16 = (x^2 - x/4 + 1/8)e^(4x)#

Paisley Johnson · Answer

The answer is #=e^(4x)(x^2/4-x/8+1/32)+C#

Integration by parts is

#intuv'dx=uv-intu'vdx#

Here, we have

#u=x^2#, #=>#, #u'=2x#

#v'=e^(4x)#, #=>#, #v=e^(4x)/4#

Therefore,

#intx^2e^(4x)dx=(x^2e^(4x))/4-int2x*e^(4x)dx/4#

#=(x^2e^(4x))/4-1/2intx*e^(4x)dx#

We apply the integration by parts a second time to find #intx*e^(4x)dx#

#u=x#, #=>#, #u'=1#

#v'=e^(4x)#, #=>#, #v=e^(4x)/4#

Therefore,

#intx*e^(4x)dx=x/4e^(4x)-inte^(4x)dx/4#

#=x/4e^(4x)-e^(4x)/16#

Putting the results together,

#intx^2e^(4x)dx=(x^2e^(4x))/4-1/2(x/4e^(4x)-e^(4x)/16)#

#=e^(4x)(x^2/4-x/8+1/32)+C#

Charles Anctil · Answer

undefined To integrate $ \int x^2e^{4x} $ by integration by parts, use the formula:

$$ \int u \, dv = uv - \int v \, du $$

Choose $ u = x^2 $ and $ dv = e^{4x} \, dx $.

Then, find $ du $ and $ v $.

$$ du = 2x \, dx $$
$$ v = \frac{1}{4}e^{4x} $$

Now, apply the integration by parts formula:

$$ \int x^2e^{4x} \, dx = \frac{1}{4}x^2e^{4x} - \frac{1}{2} \int xe^{4x} \, dx $$

Now, integrate $ \int xe^{4x} $ using integration by parts again:

Choose $ u = x $ and $ dv = e^{4x} \, dx $.

Find $ du $ and $ v $.

$$ du = dx $$
$$ v = \frac{1}{4}e^{4x} $$

Apply the integration by parts formula again:

$$ \int xe^{4x} \, dx = \frac{1}{4}xe^{4x} - \frac{1}{4} \int e^{4x} \, dx $$

Integrate $ \int e^{4x} \, dx $ to get $ \frac{1}{16}e^{4x} $.

Now, substitute the values back into the original integral:

$$ \int x^2e^{4x} \, dx = \frac{1}{4}x^2e^{4x} - \frac{1}{2} \left( \frac{1}{4}xe^{4x} - \frac{1}{16}e^{4x} \right) $$

$$ = \frac{1}{4}x^2e^{4x} - \frac{1}{8}xe^{4x} + \frac{1}{32}e^{4x} + C $$

Where $ C $ is the constant of integration.