What is the integral of 5x^3(ln(x)-2) ?

The integral of 5x^3(ln(x)-2) is 5(1/4 x^4ln(x)-(x^4}{16}-\frac{x^4)/2)+C

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integral of 5x^3(ln(x)-2)

\int 5 x^{3} (ln (x) - 2) d x

5 (\frac{1}{4} x^{4} ln (x) - \frac{x ^{4}}{16} - \frac{x ^{4}}{2}) + C

Solution steps

\int 5 x^{3} (ln (x) - 2) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 5 \cdot \int x^{3} (ln (x) - 2) d x

Expand

x^{3} (ln (x) - 2) : x^{3} ln (x) - 2 x^{3}

= 5 \cdot \int x^{3} ln (x) - 2 x^{3} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 5 (\int x^{3} ln (x) d x - \int 2 x^{3} d x)

\int x^{3} ln (x) d x = \frac{1}{4} x^{4} ln (x) - \frac{x ^{4}}{16}

\int 2 x^{3} d x = \frac{x ^{4}}{2}

= 5 (\frac{1}{4} x^{4} ln (x) - \frac{x ^{4}}{16} - \frac{x ^{4}}{2})

Add a constant to the solution

= 5 (\frac{1}{4} x^{4} ln (x) - \frac{x ^{4}}{16} - \frac{x ^{4}}{2}) + C

x \to \infty lim (\frac{30}{1 + x ^{10}})

\int \frac{cos ( 2 x ) - 1}{cos ( 2 x ) + 1} d x

\int cos (2 π x) d x

\frac{d}{d x} ((x^{5} - 9 x)^{2})

f (x) = 4 x^{3} - 5 x + 1

What is the integral of 5x^3(ln(x)-2) ?
The integral of 5x^3(ln(x)-2) is 5(1/4 x^4ln(x)-(x^4}{16}-\frac{x^4)/2)+C