2. (6 points) use integration by parts to find the indefinite integral. ∫x³ln(x²) dx

Explanation:

Step1: Simplify the integrand

First, use the property $\ln(x^{2}) = 2\ln(x)$. So the integral becomes $\int x^{3}\ln(x^{2})dx=\int 2x^{3}\ln(x)dx = 2\int x^{3}\ln(x)dx$.

Step2: Apply integration - by - parts formula

The integration - by - parts formula is $\int u dv=uv-\int v du$. Let $u = \ln(x)$ and $dv=x^{3}dx$. Then $du=\frac{1}{x}dx$ and $v=\frac{1}{4}x^{4}$.

Step3: Substitute into the formula

$2\int x^{3}\ln(x)dx=2\left[\ln(x)\cdot\frac{1}{4}x^{4}-\int\frac{1}{4}x^{4}\cdot\frac{1}{x}dx\right]$.

Step4: Simplify the new integral

$2\left[\frac{1}{4}x^{4}\ln(x)-\frac{1}{4}\int x^{3}dx\right]$.

Step5: Integrate $x^{3}$

Since $\int x^{3}dx=\frac{1}{4}x^{4}+C$, we have $2\left[\frac{1}{4}x^{4}\ln(x)-\frac{1}{4}\cdot\frac{1}{4}x^{4}+C\right]$.

Step6: Simplify the expression

$2\left[\frac{1}{4}x^{4}\ln(x)-\frac{1}{16}x^{4}+C\right]=\frac{1}{2}x^{4}\ln(x)-\frac{1}{8}x^{4}+C$.

Answer:

$\frac{1}{2}x^{4}\ln(x)-\frac{1}{8}x^{4}+C$