Evaluate ∫ 5 4 x 3 ln ⁡ ( x ) 3 d x.

Rewrite as&nbsp;∫5x3(3ln(x))4dx.∫5x3(3ln(x))4dxSince&nbsp;5⋅34&nbsp;is constant with respect to&nbsp;x, move&nbsp;5⋅34&nbsp;out of the integral.5⋅34∫x3(ln(x))dxMultiply&nbsp;5&nbsp;by&nbsp;3.154∫x3(ln(x))dxIntegrate by parts using the formula&nbsp;∫udv=uv-∫vdu, where&nbsp;u=ln(x)&nbsp;and&nbsp;dv=x3.154(ln(x)(14x4)-∫14x41xdx)Simplify.154(ln(x)x44-∫x34dx)Since&nbsp;14&nbsp;is constant with respect to&nbsp;x, move&nbsp;14&nbsp;out of the integral.154(ln(x)x44-(14∫x3dx))By the Power Rule, the integral of&nbsp;x3&nbsp;with respect to&nbsp;x&nbsp;is&nbsp;14x4.154(ln(x)x44-14(14x4+C))Simplify the answer.154(14ln(x)x4-116x4)+C

Step by step solution to "Evaluate ∫ 5 4 x 3 ln ⁡..."

Brody Collins

Answered question

2022-05-12

Evaluate

\int \frac{5}{4} x^{3} \ln (x)^{3} d x

Answer & Explanation

Sage Guerra

Beginner2022-05-13Added 10 answers

Rewrite as $\int \frac{5 x^{3} (3 \ln (x))}{4} d x$ .

$\int \frac{5 x^{3} (3 \ln (x))}{4} d x$

Since $\frac{5 \cdot 3}{4}$ is constant with respect to $x$ , move $\frac{5 \cdot 3}{4}$ out of the integral.

$\frac{5 \cdot 3}{4} \int x^{3} (\ln (x)) d x$

Multiply $5$ by $3$ .

$\frac{15}{4} \int x^{3} (\ln (x)) d x$

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = \ln (x)$ and $d v = x^{3}$ .

$\frac{15}{4} (\ln (x) (\frac{1}{4} x^{4}) - \int \frac{1}{4} x^{4} \frac{1}{x} d x)$

Simplify.

$\frac{15}{4} (\frac{\ln (x) x^{4}}{4} - \int \frac{x^{3}}{4} d x)$

Since $\frac{1}{4}$ is constant with respect to $x$ , move $\frac{1}{4}$ out of the integral.

$\frac{15}{4} (\frac{\ln (x) x^{4}}{4} - (\frac{1}{4} \int x^{3} d x))$

By the Power Rule, the integral of $x^{3}$ with respect to $x$ is $\frac{1}{4} x^{4}$ .

$\frac{15}{4} (\frac{\ln (x) x^{4}}{4} - \frac{1}{4} (\frac{1}{4} x^{4} + C))$

Simplify the answer.

$\frac{15}{4} (\frac{1}{4} \ln (x) x^{4} - \frac{1}{16} x^{4}) + C$

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