How do you integrate #int x e^ sqrtx dx # using integration by parts?

Question

How do you integrate #int x  e^ sqrtx   dx  # using integration by parts?

Scarlett Allison · Accepted Answer

#intxe^sqrt(x)dx = 2e^sqrt(x)(x^(3/2)-3x+6sqrt(x)-6)+C#

First, we use substitution.

Let #t = sqrt(x) => dt = 1/(2sqrt(x))dx# and #t^3 = x^(3/2)#

Then, substituting, we have

#intxe^sqrt(x)dx = 2intx^(3/2)e^sqrt(x)1/(2sqrt(x))dx = 2intt^3e^tdt#

Next, we apply integration by parts three times, using the formula

#intudv = uv - intvdu#

Integration by Parts 1:

Let #u = t^3# and #dv = e^tdt# Then #du = 3t^2dt# and #v = e^t#

Applying the formula: #2intt^3e^tdt = 2(t^3e^t - 3intt^2e^tdt)#

Integration by Parts 2:

Focusing on the remaining integral, let #u = t^2# and #dv = e^tdt# Then #du = 2tdt# and #v = e^t#

Applying the formula: #intt^2e^tdt = t^2e^t - 2intte^tdt#

Integration by Parts 3:

Again, focusing on the remaining integral, let #u = t# and #dv=e^tdt# Then #du = dt# and #v = e^t#

Applying the formula: #intte^tdt = te^t-inte^tdt = te^t - e^t + C#

Substituting our result into the second integration by parts step, we obtain

#intt^2e^tdt = t^2e^t-2(te^t-e^t) + C = e^t(t^2-2t+2)+C#

Substituting this into the first integration by parts step, we obtain

#2intt^3e^tdt = 2[t^3e^t-3e^t(t^2-2t+2)]+C#

#=2e^t(t^3-3t^2+6t-6)+C#

Finally, we substitute #t=sqrt(x)# back in to obtain our final result:

#intxe^sqrt(x)dx = 2e^sqrt(x)(x^(3/2)-3x+6sqrt(x)-6)+C#

Scarlett Allison · Answer

undefined To integrate $ \int x e^{\sqrt{x}} \, dx $ using integration by parts, we can choose $ u = x $ and $ dv = e^{\sqrt{x}} \, dx $. Then, we differentiate $ u $ to get $ du = dx $ and integrate $ dv $ to get $ v = 2e^{\sqrt{x}} $.

Applying the integration by parts formula $ \int u \, dv = uv - \int v \, du $, we get:

$$ \int x e^{\sqrt{x}} \, dx = x \cdot 2e^{\sqrt{x}} - \int 2e^{\sqrt{x}} \, dx $$

$$ = 2xe^{\sqrt{x}} - 2\int e^{\sqrt{x}} \, dx $$

The integral $ \int e^{\sqrt{x}} \, dx $ can be evaluated by making the substitution $ u = \sqrt{x} $. This gives us:

$$ \int e^{\sqrt{x}} \, dx = 2\int e^u \, du $$

$$ = 2e^u + C = 2e^{\sqrt{x}} + C $$

Therefore,

$$ \int x e^{\sqrt{x}} \, dx = 2xe^{\sqrt{x}} - 4e^{\sqrt{x}} + C $$

where $ C $ is the constant of integration.