Math / Calculus

Question1) Use integration by parts or American method to evaluate: b) $\int x^{3} \cdot \ln x d x$

Studdy Solution

STEP 1

1. We are given the integral $\int x^3 \cdot \ln x \, dx$ .
2. We will use the integration by parts method to evaluate the integral.
3. Integration by parts formula: $\int u \, dv = uv - \int v \, du$ .

STEP 2

1. Identify $u$ and $dv$ for integration by parts.
2. Differentiate $u$ to find $du$ and integrate $dv$ to find $v$ .
3. Apply the integration by parts formula.
4. Simplify the resulting expression.
5. Evaluate the remaining integral, if necessary.

STEP 3

Identify $u$ and $dv$ :
Let $u = \ln x$ and $dv = x^3 \, dx$ .

STEP 4

Differentiate $u$ to find $du$ and integrate $dv$ to find $v$ :
$du = \frac{1}{x} \, dx$ $v = \int x^3 \, dx = \frac{x^4}{4}$

STEP 5

Apply the integration by parts formula:
$\int x^3 \cdot \ln x \, dx = uv - \int v \, du$
Substitute the expressions for $u$ , $v$ , and $du$ :
$= \ln x \cdot \frac{x^4}{4} - \int \frac{x^4}{4} \cdot \frac{1}{x} \, dx$

STEP 6

Simplify the expression:
$= \frac{x^4 \ln x}{4} - \frac{1}{4} \int x^3 \, dx$

STEP 7

Evaluate the remaining integral:
$\int x^3 \, dx = \frac{x^4}{4}$
Substitute back:
$= \frac{x^4 \ln x}{4} - \frac{1}{4} \cdot \frac{x^4}{4}$

STEP 8

Simplify the final expression:
$= \frac{x^4 \ln x}{4} - \frac{x^4}{16} + C$
where $C$ is the constant of integration.
The evaluated integral is:
$\boxed{\frac{x^4 \ln x}{4} - \frac{x^4}{16} + C}$

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