Home > Calculus > Integration by Parts

How do you integrate #ln(x) x^ (3/2) dx#?

Answer 1

Elijah Abram

#intln(x)x^(3/2)dx=2/5x^(5/2)(ln(x)-2/5)+C#

For this problem, we will use integration by parts:

Let #u = ln(x)# and #dv = x^(3/2)dx#

Then #du = 1/xdx# and #v = 2/5x^(5/2)#

Applying the formula #intudv = uv - intvdu# gives us

#intln(x)x^(3/2)dx = 2/5x^(5/2)ln(x)-int2/5x^(5/2)*1/xdx#

#=2/5x^(5/2)ln(x)-2/5intx^(3/2)dx#

#=2/5x^(5/2)ln(x) - 2/5(2/5x^(5/2))+C#

#=2/5x^(5/2)(ln(x)-2/5)+C#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Elijah Abram

To integrate (\ln(x) \cdot x^{3/2} , dx), you can use integration by parts. Let (u = \ln(x)) and (dv = x^{3/2} , dx). Then, (du = \frac{1}{x} , dx) and (v = \frac{2}{5} x^{5/2}).

Applying integration by parts formula:

[ \int u , dv = uv - \int v , du ]

yields:

[ \int \ln(x) \cdot x^{3/2} , dx = \frac{2}{5} x^{5/2} \ln(x) - \int \frac{2}{5} x^{5/2} \cdot \frac{1}{x} , dx ]

Simplify and integrate:

[ \int \ln(x) \cdot x^{3/2} , dx = \frac{2}{5} x^{5/2} \ln(x) - \frac{2}{5} \int x^{3/2} , dx ]

[ = \frac{2}{5} x^{5/2} \ln(x) - \frac{2}{5} \cdot \frac{2}{5} x^{5/2} + C ]

[ = \frac{2}{5} x^{5/2} \left( \ln(x) - \frac{2}{5} \right) + C ]

where (C) is the constant of integration.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.