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

The integral of 2x^3ln(x^2) is x^4ln(x)-(x^4)/4+C

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

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

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

Solution steps

\int 2 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

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

Apply Integration By Parts

: \frac{1}{4} x^{4} ln (x^{2}) - \int \frac{x ^{3}}{2} d x

= 2 (\frac{1}{4} x^{4} ln (x^{2}) - \int \frac{x ^{3}}{2} d x)

\int \frac{x ^{3}}{2} d x = \frac{x ^{4}}{8}

= 2 (\frac{1}{4} x^{4} ln (x^{2}) - \frac{x ^{4}}{8})

Simplify

2 (\frac{1}{4} x^{4} ln (x^{2}) - \frac{x ^{4}}{8}) : x^{4} ln (x) - \frac{x ^{4}}{4}

= x^{4} ln (x) - \frac{x ^{4}}{4}

Add a constant to the solution

= x^{4} ln (x) - \frac{x ^{4}}{4} + C

y = 3 + 2 x, y = \frac{9 + x}{3}

\frac{d}{d x} (a^{f (x) (x^{2})})

\frac{\partial}{\partial x} (e^{y} (y^{2} - x^{2}))

(tan (x))^{^{'}}

y = - 2 x^{3}, (- 2, 16)

What is the integral of 2x^3ln(x^2) ?
The integral of 2x^3ln(x^2) is x^4ln(x)-(x^4)/4+C