How do you find the integral #ln(x) x^(3/2) dx#?

Question

Isabella Ackerman · Accepted Answer

To carry out component integration: #uv = intvdu# I would pick a #u# that gives me an easy time differentiating, and a #dv# that I would easily be able to integrate multiple times if necessary. If the same integral comes back, or variables aren't going away, either it's cyclic or you should switch your #u# and #dv#. Let:
#u = lnx#
#du = 1/xdx#
#dv = x^(3/2)dx#
#v = 2/5x^(5/2)# #=> 2/5x^(5/2)lnx - int2/5(x^(5/2)/x)dx# That's not too bad, nice. #= 2/5x^(5/2)lnx - int2/5x^(3/2)dx# #= 2/5x^(5/2)lnx - 4/25x^(5/2) + C# #= 10/25x^(5/2)lnx - 4/25x^(5/2) + C# #= x^(5/2)[10/25lnx - 4/25] + C# #= 2/25x^(5/2)[5lnx - 2] + C#

Samantha Adkison · Answer

undefined To find the integral $\int \ln(x) x^{3/2} \, dx$, we can use integration by parts. Let $u = \ln(x)$ and $dv = x^{3/2} \, dx$. Then, we have $du = \frac{1}{x} \, dx$ and $v = \frac{2}{5}x^{5/2}$.

Applying the integration by parts formula:

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

we get:

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

Simplify the integral:

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

Now, integrate $\int x^{3/2} \, dx$ to get:

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

So, the integral of $\ln(x) x^{3/2} \, dx$ is:

$$
\frac{2}{5}x^{5/2} \ln(x) - \frac{4}{25}x^{5/2} + C
$$